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LIBRARY OF CONGRESS. I 



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J UNITED STATES OP AMERICA. 



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KEY 



ROBINSON'S 



NEW GEOMETRY AID TRIGONOMETRY, 



CONIC SECTIONS AND ANALYTICAL GEOMETRY. 



SOME ADDITIONAL ASTRONOMICAL PROBLEMS. 



DESIGNED FOE TEACHERS AND STUDENTS, 



_ Hac 

NEW YORK: 
PUBLISHED BY IVISON, PHINNEY & CO., 

48 & 50 WALKER STREET. 
CHICAGO : S. C. GRIGGS & CO., 39 & 41 LAKE ST. 

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ROBINSON'S 

*f j§§atfea»atic^ 

27^e most Complete, most Practical, and most Scientific Series of 
Mathematical Test-Books ever issued in this country. 

(33ST TWENTY VOLUMES.) 
— ii 1 1 ^ 

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IV. Robinson's Progressive Practical Arithmetic, 56 
"V. Robinson's Key to Practical Arithmetic, 50 

"VI. Robinson's Progressive Higher Arithmetic, 75 

VII. Robinson's Key to Higher Arithmetic, 75 

VTIL Robinson's IsTew Elementary Algebra, ----- 75 

IX. Robinson's Key to Elementary Algebra, 75 

X. Robinson's University Algebra, - 1 25 

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a/ 7 i<& 



PREFACE. 



A Key to a Text-book on the Higher Mathematics, 
if not a new creation, is by no means a common thing. 
And it is a question undecided in the minds of many, 
whether a Key to any mathematical work is an aid, or 
a hindrance to the teacher. 

The value of a Key depends upon the use to which 
it is applied. Class, and school-room duties so fully oc- 
cupy the time of a majority of teachers, as to make a 
Key a great convenience, if not a necessity. And teach- 
ers of limited acquirements, as well as private students, 
will often find a Key of great service. 

A Key should never be used to supersede investiga- 
tion and labor, but to give direction to study, and make 
labor more effective. It should lessen the mechanical 
labor of teaching, by showing how to study, and how 
to teach, by giving the best forms of analysis, the best 
arrangement of the work, new and improved methods 
of operations, and often by developing methods of solu- 



i v PREFACE. 

tion too elaborate to find a place in the Text, thus 
giving to those who use a Key, more extended and 
enlarged views of the applications of Mathematical 
Science. 

In the preparation of the present work, great care 
has been taken to give full and comprehensive solutions 
of the examples and problems in Kobinson's Plane and 
Solid Geometry, Plane and Spherical Trigonometry, and 
Analytical Geometry, and by means of cuts, more fully 
to explain and demonstrate the principles involved, thus 
making it a sort of commentary on the Text itself, and 
which in the hands of a good teacher may prove a val- 
uable auxiliary in teaching. 

Several pages of miscellaneous Astronomical examples 
have been added after Trigonometry. As soon as the 
New Calculus and Surveying, which are in preparation, 
are completed, a Key to those works will be also added 
to this. 

January 1st., 1862. 



KEY TO 

R O B I N S O N'S 

NEW GEOMETRY AND TRIGONOMETRY. 



PRACTICAL PROBLEMS. 

(Book V.— Page 143.) 

Ex. 1. The vertical angle being 60°, and the sum of all 
the angles being 180, therefore the sum of the other two 
angles, is 120°, and, as these angles must be equal to each 
other, each one must be 60°, and the triangle is equilateral 
as well as isosceles. Each side is therefore equal to the 
base 6. 

Ex. 2. As the two oblique angles of any right angled 
triangle equal 90°, and as one of them is 30°, the other must 
be 60°. Describe an equilateral triangle, each angle of which 
will be 60°. Divide either angle into two equal parts, by a line 
which will divide the opposite side into two equal parts, and 
the triangle into two equal triangles. The least angle, in 
either of these triangles is 30°, and the side opposite 30° will 
be half of the hypotenuse, and this is a general truth. 
Namely, That in every right angled triangle whose least 
angle is 30°, the side opposite that angle is one-half of the 
hypotenuse. 

Therefore, if in this example the least side is 12, the hy- 
potenuse is 24. 



KEY TO 



[143 



Ex. 3. Let AB, CD, represent two parallels, and AC the 
perpendicular distance between them, equal 10. Take AD, 
in your divides, double the dis- E 
tance A C, and strike an arc, cut- 
ting CD in D. Join A D, and the 
angle ADC 'will be equal to 30°, 
because ACD is a right angled C 
triangle, and its hypotenuse double the side A C (By Ex. 
2). But DAB— ADC, being alternate angles, and because 
DAB=30°, its adjacent angle is 180° -30° =150°. 

Hence the answer must be 30° and 150°. 




Ex. 4. We may use the same parallels as in (Ex. 3), to 
illustrate this example. Let AC=20, take CD=20, and join 
AD, and then will AC=CD, and ACD will be an isosceles 
triangle, and the angles ADC and CAD, equal to each other. 
Their sum must be 90°, therefore each one of them must 
be 45°. Now as CAB=90°, and CAD=45°, therefore, 
DAB=45°, and DAE=W D + 45°=I35°. 

Because AC '' +GD' ' =AD\ AC=20, 



AD =800=400 x 2. Whence AD =20 V2. A 



ns. 



Ex. 5. Let DE be one parallel, and AB a perpendicular 
to the other parallel. Take BE= 
BC and BD any distance greater 
than BC Assume AD = 15, AC= 
AE=10, and AB=8. Now, by the 
well known property of the right 
angled triangle, we have 

CB 2 =BE 2 =lQ>-&=6 2 . Whence, CB=6=BE. 

Also, 55 2 =15 2 -8 2 =23x 7=161 

Whence, DB= VT61=12.69, and DE= 18.69 ) . 

3 > Ans. 

and DC= 6.69 S 




143, 144] GEOMETRY. 7 

Ex. 6. This problem is the same as (Ex. 5,) except in 
data, therefore use the same cut as before. 

Make AB =12, AD =20, AC=AE=18. 
Now, 5S 2 =(20 2 _12 2 )=32 x 8=16 x 16. Whence, BD=16. 
CB*=BE 2 =18* -12 2 =30x6=180. 05=13.416. 

Sum, BU=29A16 ) 

DifF. DC= 2.584 \ 

The area of the triangle ADE= (29.416)6=176.496 



Ans. 



The area of the triangle ADC={ 2.584)6= 15.504 



Ans, 



Ex. 7. This example produces a right angled triangle, the 
hypotenuse is 6, and one side 2, one half of the chord. 
Therefore the other side is, 

|/6^2 2 = |/32=4|/2. Ans. 



Ex. 8. The two parallel chords being equal, they must be 
equally distant from the center, and be on opposite sides 
of the center. 

Let AB, CD, be those paralells ; 
and from the center draw OB, OD ; 
and OE, OG, each perpendicular 
to AB, and CD ; then EOG, will 
be one right line through the cen- 
ter. EG=6, therefore E0=3. 
AB=8 ; hence, EB=4 ; and in 
the right angled triangle EOB, we 
haveOS 2 =4 2 + 3 2 =25. Whence, 05=5. Ans. 




Ex. 9. This is a problem of the same kind as (Ex. 8,) 
and therefore we may use the same cut. 
Let AB, CD, be the parallel chords. Put AB=16, 



8 KEY TO [144 

CD=12, and EG=14. Place OB=OB=B, EO=x } 
OG=y. Then EG=x+y = 14, or y=U-x. 

And a 2 + 8 2 =i? 3 
Also y 2 + 6 2 =B 2 

Whence x 2 + 8 2 =y 2 + 6 2 =(l4-x) 2 + V 
x 2 +28=I¥-28x+x 2 
Or 28£ + 28=14 2 

2a + 2 =14, or x=6. Whence y=8, 
And B 2 =8* + 6 2 =100, or 5=10. -4tw. 

Ex. 10. Let # represent the perpendicular required, 
which is the perpendicular drawn from the vertex of an 
isosceles triangle to the center of its base. 

Then a 2 =15 2 - 5* =20. 10 =2(10)*, or x=10V2. Arts. 

Ex. 11. Let ABOhQ the triangle, bisect CB in E, AC 
in F. And join AE and BF, and through the point of in- 
tersection 0, draw CO, and produce it to D. 

Because CB is bisected in E, the triangles ABE and A EC 

are equal. Or the triangle A EC 
is half the triangle ABC. 

In like manner, we prove that 
the triangle BFC is half ABC, 
33 B because A C is bisected in F. 

But BOC consists of two equal triangles, BOE and EOC, 
each of which may be represented by b, and AOC consists 
of the equal triangles, A OF and FOC, each of which may 
be represented by a. Therefore BFC consists of 2b -\- a, and 
AEC consists of 2a + 5. Whence, 2b + a=2a + b. 

That is, b=a. 

Now it is obvious that, BFC is equal to three triangles, 
each equal to a, or each equal to b. 




£44] GEOMETRY. 9 

Hence, the triangle AFB must be equal to 3a, and AOB 
equal to 2a. But the triangle ABC is equal to 6a. That 
is, AOB is equal to one third of ABC. 

Or, 6a to 2a as 3 to 1. Ans. 

Ex. 12. If the diameter of a 
circle is 32, its radius is 16. 
Hence, <7£=16, CA = 16, and 
we have two right angled trian- 
gles, CGB and CD A; GB =10, 
half of one chord, and AD=A, half of the other parallel 
chord. 

Now, these two right angled triangles give us 
£S=16 2 -4 2 =20x12=240, 
C67 2 =16 2 -l0 2 =26x 6=156, 
Whence CD= V240= 15.49 + 

And (7£= ^156=12.49 + 




DifT.=D6r=3, if the parallels he the same side of the 
center, but when on opposite sides, their sum 27.98 will be 
the distance between the parallels. 




Ex. 13. Here GD=12, AB=8, BZ>=5. Conceive CA 
and DB produced until they meet, thus 
forming a triangle of which AB is the 
base. Also conceive a line drawn from 
the vertex of this triangle perpendicular 
to the base, and designate it by x. 

Now by proportional triangles, we shall have, 
x : 8 : : x + 5 : 12 
And 12^=8^ + 40. Whence, a=10. 

The area of the triangle on AB as & base, is therefore 




10 KEY TO [| 45) 146 

"JO i Q 

4#=40. The area of the trapezoid is -^—^ x 5=50, and 
the area of the whole triangle on the base CD is 90. Ans> 



Ex. U. Let ABO be the 
triangle AB=697,. AC=813, 
££=534. 
Now by (Th. 24, B. II.), we 
"d b have 

AD : DB : : 813 : 534 
271 : 178 
Whence, 178 A D= 271 DB. And AD +DB =697 

178 XD + 178 £,8=697x178 
That is, 271 DB +178 DB=697 x 178 

Or 449£Z>=697xl78 

. 697x178 

BD= 

449 
Thus we find £D=276.316, the less part. 



.4 729 

Whence AD=420.684, the greater part. 

Examples 15, 16, and 17, solutions in text-book 

Ex. 18. By referring back to (Ex. 15), we observe, that 
EG must be taken from DB. But DH= V51, and HC= ^40, 
hence if HO were taken from DH, the point C from D would 
then be (V5l— v'40). But AD=16. Whence, according to 
this condition, 

AC*=(V5i-VM\i6* 

= 91-21/2040 + 256=347-2^2040. 
Or, AC'=347-2 1/2040=347-90.332=256.668. 
Whence, AO=16.021 



_ _ } Ans. 
Area of the triangle is -} (9 /51 + 7 ^40) . 



147] 
Ex. 19. 



GEOMETRY 



11 



Let E be the position of the observer's eye, 




EF=40 feet, FH=24 
feet, ED the required 
distance which we de- 
signate by x, and DB= 
90 rods, which must - 
be reduced to feet thus 90 x 16 J, or 45 x 33. 

Now by proportional triangles EFH, EDB, we have, 

40 : 24 : : x : 45 x 33 
Or, 10 : 6 : : a : 45x33 or, 6^=45x330 

2^=45x110 
or, x— 2475 feet. Ans. 




Ex. 20. Illustrated in text-book. 

Ex. 21. Let ABO be a triangle divi- 
ded into two parts by the line DE, par- 
allel to the base AB. 

Let CDE be taken, as 1, and the other 
part ABDE, as 3. Then the whole trian- 
gle ABO will be represented by 4, the trian- 
gle ODE, being unity. 

Now by (Th. 22, B. II.), we have, 
CD* : OF : : 1 : 4 
Square root, CD : OB : : 1 : 2. 

Whence CB=2CD, or CD equals one half of OB, and 
therefore CE equals one half of OA. 

Ex. 22. Let the last triangle represent the triangle in 
this example also. That is, place ^4i?=:320, and as the 
angles adjacent to the base are 90° and 60°, the angle oppo- 
site at G must be 30°. The side opposite 90° must be 



12 key to [147—149 

double of the side opposite 30°, as was explained in (Ex. 2), 
of this key. Therefore, AC=640. 
Again, because the angle 5=90°. 

That is, 320 2 + W=640 2 

Or, ^ 2 =640 2 -320 2 =(320) 2 (2 2 ~l)=:320 a x 3. 

Whence, £(7=320 VI =554.24. Ans. 




Ex. 25. Draw AB of any indefinite length, and from 
the point A draw AC, making an angle of 32° with AB, and 
from B draw the line BC, 

making with BA an angle 
equal to 84°. Produce AB 
to D. Because the sum of 
the three angles of any trian- 
gle is equal to two right an- 
gles, the angle C of the triangle ABC, must be 
C= 180° -(32° + 84°) = 64. 
The exterior angle CBD=180°-84 o =96°, and because 
EBB is one half of it, the exterior angle EBB of the trian- 
gle AEB is 48°. But the exterior angle EBB=E+EAB= 
E+16°. 

Therefore, £ r +16°=48°, and ^=32°, one half the angle 
C, a general result. 

Ex. 26. Let AB and CD be two parallels, and AC a line 
A . _ between them ; because the two inte- 

X> ' 

rior angles between two parallels are 
equal to two right angles, therefore, 




.D 



one half of each of them, added to- 

gether, must make one right angle. That is, the angle EA C, 
and EGA together, make one right angle, therefore the third 
angle at E, of the triangle AEC, must be a right angle. 



149] 



GEOMETRY. 



13 



Ex. 27. Let ABCD be any trapezoid, and draw its 
diagonals AD and CB, intersecting in E. The two triangles 
AEB and CED are equi-angular A B 

and similar. The opposite angles 
at E are equal ; and because, AB 
and CD are parallel, the alternate 
angles EAB and EDC are equal, 
and so are the alternate angles EBA and ECD, therefore the 
two triangles AEB and CED, are equi-angular, and AB and 
CD, sides opposite the equal angles E, are homologous. 
Therefore, by (Th. 20, B. II.), Ave have 

Z5 a : CB\ as AEB is to CED. 





Ex.28. Let A BCbQ any triangle, 
the sum of its three angles is 180°. 
Take any point P within it, and 
draw PB, PC. If we designate the 
angle between AB and PB by a, and 
between AC and PC by b, then the angles at the base of the 
triangle PBC, will be less than the angles at the base of the 
triangle ABC by the angles a and b. But as the sum of the 
angles of every triangle make two right angles, therefore the 
angle BPC is greater than the angle A, by the sum of the 
two angles a and b. 



Ex. 29. Let AB and CD be 12 and 

20 respectively, and the perpendicular 
AG equal 8. Let UK represent the 
line between the parallels. Then 
HL = 1U— 12=2i. And we have, 

CE:AG=EL : Ag 




Or, 



8=2^ 



Ag. 



14 KEY TO 

Hence, 

Area of ABCD is 



[149, 150 



Ag=2±, and gG=5± 
20 + 12 



x8 =128. 



Area of ABKH is — ± — x2±= 33k 



Area of flKDC is 2 °+ 145 x 5i=94f. 




Ex. SO. The triangles -4 0D 
and ADB are similar, and there- 
fore, the sides about the angle 
A are proportional, giving the 
proportion 

AC: AD : : AD : J5. 
But AC expressed in feet is 
the product of 7956 by 5280, 
plus AB=40 feet. 

7956x5280=42007680. 
Now the above proportion becomes, 

42007720 : AD : : AD : 40 
Whence, AD 2 = 1 680308800 

Sq. root AD =40992 feet nearly, or 7 miles and 

4032 feet. 




Ex. 31. As the triangle ABD 
in the circle is to be equilateral, 
it must also be equiangular. Hence, 
the line A C from one of the angles 
to the center must bisect the an- 
gle A, and each division must be 
30°. AC=2, and the side opposite 



150] GEOMETRY. 15 

30° is half AC, therefore, CE=1, and Z>£=3. Place 
AE=x, 

Then, the right angled triangle AEQ gives x" + 1— 4. 

Whence, #~ 1^3. But AE into Di7, is the area sought. 

That is, 3^3, is the area of the triangle ABD. 




Ex. 82. Let AB—12, AC=11, 05=10. From C draw 
CD perpendicular to AB. Take H the middle point be- 
tween A and B, and draw iLP 
at right angles to AB, until it 
meets a perpendicular drawn 
from the middle point between 
A and C; thenPAvillbethe loca- 
tion of the well. Draw PG ^ 6 H D 
parallel to AB, AH=HB=6. PH is common to the two 
triangles APH and PHB, therefore, AP=PB, AP=PC 
by construction. Place PE=x and HD=y. 

Then, u4Z>=6 + y and DB=6—y. 
Now (6 + 7/) 2 + GD 2 =lT, and (6-*/) 2 + O# 2 :=l0\ 
By substitution, 24?/ = 121 -100=21, or y=i 
This value of y substituted in the preceding equation 
gives (5|) 2 + OS 2 = 100, 

Or, 25 + -Y-+o 1 t + £5 2 =100, 

Or, 210 + i + 3O5 2 =800. 

Whence, 80^=589.875, and CD=8.587. 
Secondly. In the right angled triangles AHP and CPG, 
we have, 

6 2 +x 2 =B\ B=AP=PC=PB. 
And (8.587 -xy + (iy=B\ 

From these two equations, we obtain x, and afterwards R. 



16 



KEY TO 



[150, 131 



Ex. 03, The required perpendicular will obviously bisect 
the base of the triangle, and divide the figure into two equal 
right angled triangles. Eepresent the common perpendicular 
by x. Then x* + 36=400, or, o; 2 =364, a; =19.07. 

Area sought =(19.07)6. 

Ex. 85. This problem is a supplement to (Ex. 34), which 
is explained in the text-book. 

In (Ex. 34), we have the two sides of a right angled trian- 
gle (27.3) and (35.76). This problem demands the hypote- 
euse to that triangle, and its division into parts in the ratio 
of (27.3) to (35.76). (Th. 24, B. II.) 

Place a=27.3, and 5=35.76, and x + y= the required 

(i) 



hypotenuse, then 



But, x : y : : a : b. Or, bx 

Multiply (1) by b, then 

bx + by=bVa*^F z 



x-\-y=zVa* 
ay 



¥ 



(2) 



That is, ay + by^bVcf + b* 



And 



x=- 



bVtf + b* 

a + b 
aVcf + V 



a + b 



Ex. 36. If 6 feet makes a descent of 1 foot, what dis- 
tance will be required for a descent of 4 feet. Eepresent 
that distance by x. 

Then, 6 : 1 : : x : 4. Or, <s=24. Ans. 




Ex. 37. Let ABO represent the tri- 
angle, and BE a side of the inscribed 
square. 

Place AB=x, Z>i7=12=a. 
Now by proportional triangles we have, 
a(x + a) 
x 



a : :x + a:BO. Whence BC- 



101] GEOMETRY. 17 

Also, AB 2 +BC?=A(T 

That is, (x + ay+-^x + ay=35 2 

X 

Or, xXx + aY + aXx + ay^'x 2 
That is, O 2 + a 2 ) (x + a) 2 = 3§V 

Or, (a 2 + a 2 ) (a; 2 + 2aa + a 2 ) = (35) V 
Whence (x* + a 2 ) 2 + 2az(V + a 2 ) = (35) V. 
Place 2/=a 2 + a 2 . Then, ?/ 2 + 2aa.y = (35) V. 

y' + 2aa^ + aV=(a a + (35)V 
= 1369x 2 . 
Square root y+ax=37x, 

y=25x. 
But 2/=a 2 + 144. Therefore, x 2 + 144=25x 

x~-25x=-144. 
From this equation we find, #=16, or 9. 

That is, 
AB=16 or 9. But BB=12, whence AB=28 or 21. Ans. 

To find 5(7 we have 30=J£±4J2Q*^ =2L 

x 16 

Or, ??£±!2> =2 a 

y 

Ex. 38. Taking the triangle as in the preceding Example, 
we observe that the least hypotenuse in a triangle containing 
the same square, must require AB=BE, or AB—BC. 

If AB=BE, AB must be equal to 2BE, because Z>.#= 
BE. Therefore AB=z2a, and BC=2a. Each side double 
the side of the square. 



BOOK VII. 

PRACTICAL PROBLEMS. 



(Page 229.) 

Ex. 1. The result sought is }ttD\ |tt=0.5236, and 
D = 12. Whence, £7^ = (1728x0.5236) = 904.78. A?is. 

Ex. 2. The result required in this problem, is expressed 
by lnH%3B-H). (See Geometry, page 229). 

When applied to this problem, J2 = 3, B=6, tt= 3.1416. 
Whence, 

\ttH 2 (3B-H) =(1.0472)9 x 15= 141.372 cubic in. Ans. 

Ex. 3. To solve this problem, we place 4nB 2 = 68, 2B 
being the value sought. Whence, 

^Sr 4652 - Ans - 

Ex. 4. In this problem the result sought is 2nBH y but in 
the preceding Example, the value of 2B was found to be 
4.652 feet. H=2 feet. 2tt = 6.2832. 

Therefore, the result sought is the product of the factors, 
(4.652) (6.2832) = 29.229 + square feet. Ans. 

Ex. 5. For the solidity of any sphere, we have ^D% 
irr^O.5236, and in this problem Z>=4.652. Whence, 

}nD* = (4.632) 3 (0.5236)=52.72. Ans. 



230] GEOMETRY. 19 

For the solidity of the segment, we have the general for- 
mula, ±7tH 2 (3B—II), and for this example we have 11=2 
feet, and .#=2.326. Whence, 3B-H=4.978, and the 
solidity sought is the product of the three factors, 

(1.0472)4(4.978), which is 20.852. Arts. 

Ex. 6» The general expression for the solidity of a seg- 
ment of a sphere having two bases, is 

B' and B" being the radii of the bases, and H the perpen- 
dicular distance between them. 

In this Example the radius of the sphere is 10, and H=2. 
(JR'Y and (#") 2 must be computed. 
For the greater segment, we have, 

(#7 = 100-9, and (#") 2 = 100-25. 
Whence, 

(0.5236)(8) -f(3.1416)(166) = 525.7 cubic feet. Arts. 
For the smaller segments, we have, (B'f =100—25=75. 
and (i2") 2 = 100-49 = 51. 

Whence, 

4.1888+(3.1416)(126)=400.0304 cubic feet. Arts. 

Ex. 7. Let the circle AFB represent a central section of 
the sphere, FE the diameter of the seg- ^ 

ment required. Let EE=8, AD=4, and 
AO=EC=B; then in the triangle ODE, 
BC=B-4, 
And, (i^-4) 2 + 8 2 =# 2 . Whence, B=10. 
Now for the solidity of the segment, we 
have 

lnH*(3B-H) = (1.0472)16(30-4) =435.6352. Ans. 




20 



KEY TO GEOMETRY 



[230 



6 


\ T3 


1 \. 

/ \ 
1 \. 
1 \, 





\ 
\ 

I 

- ;E 



H 



Ex. 8. Let ^4j5 and J927 be the diameter of the segment, 
^-----•^ and they are parallel, and their 

distance asunder being 9 inches, 
AG=7, and DH=9, and AG and 
DG= each to R. Place CH=x, 
and the two right angled triangles 
A GC and DCH, give 
^ "'" T + (9-xy=B% F+af=B*. 

Whence, 49 + 9 2 -18£ + £ 2 = 9 2 + x 2 , or 18x=49, x=2{%. 

■vr z? 2 Qi , / 49 V 26244+2401 28645 

.#=9.4027. ^rcs. 
For the value of the segment, we have, 

.5236.(729) + 9 -( 8 - 1416 ) (8 i + 49) _ 2 219.5 + 

A 

Ex. 9* In this example, the height of the segment is 6, 
and the solution is 

.5236(216) + I M^ ] ^(300+ 144)^4297.7. 



Ex. 10. The contents of the segments will be 
itt(2 3 ) +^(100+ 36)=431.45 cu. mi. 



To find the diameter, 
Let CA=x, GG=tj 

Then, CH=y + 2. Hence, 
a 2 =s/ 2 + 100 (1) 

a 2 =(2/ + 2) 2 + 36 (2) 
Whence we have 

y=15. 




And from (1), 



a;=V225+ 100= 18.027, radius. 



BOOK V III. 

APPLICATION OF ALGEBRA TO GEOMETRY. 



PROBLEM VI. 



In a right angled triangle, having given the base, and 
the sum of the perpendicular and hypotenuse to find these 
two sides. 

Let ABC be the triangle, AB—b, and 
AC+OB=S, and BC=x. 

Whence, AC=S-x. 

But AB 2 + BC 2 =AC 2 

That is, b t +a?=(S-xy=&—2Sx+a? 
28x=8*-V 




x= 



8 2 -V 

28 



:GB, AC=8- 



&-b* &+b % 



28 



28 ' 



PROBLE 



VII 



Given the base and altitude of a triangle, to divide it into 
three equal parts, by lines parallel to the base. 

Let A BC represent the A. Con- 
ceive a perpendicular let drop from 
C to the base AB, and represent 
it by b. Put 2a=AB. Then 
ab= the area of the triangle. 

Let x be the distance from O to 
FD ; then by (Th. 22, Bk. It), A 
x* : h 1 : : \ab : ab 

Whence, x : b : : 1 : ^/S. Or, x— — . 

\/6 




22 



KEY TO 



[236 



If x represents the distance from G to GE, then 
x~ : b 2 : : ?a6 : ab. 



Or, 



a? : 6 : : y'2 : ^/3. 



V 3 ' 



We perceive by this that the divisions of the perpendicu- 
lar are independent of the base, and that we may divide the 
triangle into any required number of parts, m, n, p, etc., 
equal or unequal. 



PROBLEM VIII. 

In any equilateral triangle, given the lengths of three per- 
pendiculars drawn from any point within to the sides, to de- 
termine the sides. 

Let ABG be an equilateral tri- 
angle, and because CD is drawn 
perpendicular to the base, it bi- 
sects the base. Place AD—x. 
Then AB, BC, AC each equal 2a. 
Take PO = a, PG=b, PH=c. 
The area of the triangle ABC 
equals {CD)x. But in the right angled triangle ADC, we 
have, CD~ + x'=4x*. Whence, CD=x f /3, and the area of 
the triangle ABC, therefore is x 2 */S. 

Again, the area of the triangle APB, is (PG x ~~), 

or bx. Area BPC=ax and APC—cx, and the sum of these 
three interior triangles, equals the triangle ABC 

That is, x 2 I 7 3 = ax-\-bx-\-cx 




G D 



Whence, 



2x-. 



2(a + b + c) 



the length of a side. 



236] 



GEOMETRY. 



23 



PROBLEM IX. 



In a rigid angled triangle, having given the base (3), and 
the difference between the hypotenuse and perpendicular (1) 
to find the sides. 

Place AB=3, AC-OB=l. 

Make CB—x, then AC=l+x. 

And 9-\-x 2 = l + 2x + x\ 
Whence, 



2x=8,x=4. 



Then, ^0=5 A 




PROBLEM X 



In a right angled triangle, having given then hypotenuse 
(5), and the difference between the base and perpendicular 
(1), to find the sides. c 

Place AC=5, CB-AB=1, AB=x. 

Then CB=l + x. 

Nowx 2 + (l + ^) 2 =25. 
2x 2 + 2x = 24. 
Whence, x=3, BC—4. Ans. 




PROBLEM XI. 



Having given the area of a rectangle inscribed in a given 
triangle to determine the side of the rectangle. 

When we say that a triangle is 
given, we mean that the base and 
perpendicular are given. 

Let ABG be the triangle, AB 



= b, OD=p, CI- 

p—x. 



then ID 




D H 



24 key TO [336 

By proportional triangles, we have, 

CI : EF : : CD : AB 

Tint* 

That is, x : EF : : p : b. EF=—. 

P 

By the problem — (p— x)=a. The symbol a being the 
given area. 



Whence, x*—jpx= — -± > -. x—\p±_y ip 2 — °j~. 



PROBLEM XII. 

In a triangle having given the ratio of the tivo sides, to- 
gether with both the segments of the base, made by a per- 
pendicular from the vertical angle, to determine the sides of 
the triangle. 

Let ACB be the A, (see last figure.) AD— a, BD=b, 
and CD = x. Then A C= Va'+x 2 , and CB = V¥+ti\ 

The ratio of AC to CB is given, and let that ratio be as 1 
to r ; then 





Vct + rt : W+x* : 


; : 1 


: r. 


Whence, 


a 2 + cc 2 : tf+x' : 


: : 1 


: r\ 


Or, 


& 2 -f- x* = aV 2 + r*SD\ 






Or, 


a aV 2 -Z> 2 

x =—- . 

1-r 2 







But ^ C— tV+cc 2 , and as or is now known, AC is known. 



PROBLEM XIII, 



.Zift em?/ triangle having given the base, the sum of the other 
two sides, and the length of a line draivn from the ver- 
tical angle to the middle of the base, to find the sides of the 
triangle. 



230] GEOMETRY 

Let ADE be the A. Suppose 
G to be the middle of the base. 

Put AG— a, DO or CE=b, 
AE=x, DA+AE=c; then DA 
=c—x. 

Now by (Th. 42, B. I.), we have, 
(DA) 2 + (AEy=2(AC) 



25 




2(1)0)* 



That is, c 2 — 2cx+2x 2 =2a 2 + 2& 2 . 
Or, 4x 2 -4cx + c 2 =4a 2 + 4b 2 -c\ 

2x- 



. c =V4a 2 + 4b 2 —c 2 . 
Whence x becomes known, and consequently the sides be- 
come known. 



PROBLEM XIV. 

To determine a right angled triangle, having given the 
length of tivo lines drawn from the acute angles to the mid- 
dle of the opposite sides. 

Let ABC be the triangle. Bisect AB in 
E, OB in D, and join CE=a, and AD=b. 



Place DB=x, and BE= 



V- 



Now in the two right angled triangles 
ABD, and OBE, we have, 

4if + x 2 =b 2 (1) 

4x 2 +tf=a 2 (2) 

Sum 5(x 2 + if) = (a 2 + V) 

Or, aMY=i(a'+&") (3) 

But 4x 2 + if=a 2 (4) 

Sub. (3) from (4), and 3x 2 = ±a 2 - }b 2 . 
Whence, x= ( T *-a 2 — T \b 2 ) 

Also, ^Kt^-tVO 



i 

J V2 




26 



KEY TO 



[23© 



PROBLEM XV 




To determine a right angled triangle having given the per- 
imeter and the radius of the inscribed circle. 

Let ABO be the triangle, draw 
AO, CO to the centre of the circle,, 
and it is obvious that the two right 
angled triangles AEO, ADO, are 
equal to each other. Also, CFO= 
CEO, and BEOF is a square, each 
side equal to the radius of the circle. 

Place EB=BF=r, AE=AE=x, and CF=CD=y. De- 
signate the given perimeter by 2p. 
Then 2z+2y + 2r=2p 

Or, x+y+r=p 

Or, x + y=(p—r) (1) 

Also, by the right angled triangle, we find, 

(x+ry + (y+ry=(x+yy (2) 

Or, x 2 + 2rx + r 1 + y" + 2ry -f r 2 = cc 2 -f 2xy + y\ 
Keducing and, 2r (x + y) + 2r 2 = 2xy. 
Or, r(p— r)-\-r*=xy. 

"Whence, xy=rp (3) 

Equations (1) and (3) will readily give the values of x and 
y } which solves the problem. 



PROBLEM XVI. 



To determine a triangle, having given the base, the per- 
pendicular, and the ratio of the two sides. 



237] 



GEOMETRY. 



27 



Let ABC be the A. AB=b, 
CD=a, BB=x. Then 



AC=V(b-xy + a\ 
CB= VaT + x 1 . 
Let the given ratio of the 



sides be as m to 



then 




Vib-xy + a? : Va?-\-x 



m 




This proportion will give the value of x } then AC and CB 
will be known. 

PROBLEM XVII. 

To determine a right angled triangle, having given the 
hypotenuse, and a side of its inscribed square. 

Let ABC be the triangle, and BE— a, 
a side of the inscribed square. Place 
CB—x, and AF—y. 
Also, let AC=h. 

Then, (x + ay + (y + ay=h\ (1) 
And x : a : : a : y, (2) 

Expanding (1), transposing, etc., gives 

a?+y*+2a(x+y)=h 2 —2a\ (3) 

From (2) 2xy = 2a\ (4) 

Sum of (3) and (4) x* + 2xz/ + ?/ 2 + 2a(a + y) = h\ (5) 
Or, (g+y)'+2q(a H-y)=y . (6) 

Whence, cc+t/=:— a±|/A 2 + ^. (7) 

Equations (4) and (7) will give the value of x and y, and 
solve the problem. 

PROBLEM XVIII. 

To determine the radii of three equal circles inscribed in 
a given circle, touching each other, and each touching the 
circumference of the given circle. 

Every circle consists of 360°, and therefore if three circles 



28 



KEY TO 



[say 




are to be placed in a given circle, each one of them must 
occupy a section of 120°, one third of 360°. 

Let ACB be one of those sectors, 
and it is obvious that the centre of 
one of the interior circles must be on 
the radius CD of the larger circle, CD 
dividing the angle ACB into two 
equal parts. 

It is also obvious that the interior 
circle which is placed in the sector ACB, must touch the 
given circle in the point D, and touch the radii CA, OB, in 
the points t, t, equidistant from C, and so situated that per- 
pendiculars Ot, Ot, shall be equal to OB. 

Place OD—Ot—x, then x represents the radius of one of 
the required circles, and let R represent CD, the radius 
of the given circle. 

In the right angled triangles OtC the angle tCO=60°, 

angle 2=90°, therefore the angle t 00= 30°, CO=R— a?;and 

in every right angled triangle, when one of the acute angles 

is 30°, the side opposite to that angle is one half the hypote- 

R— x 
nuse, therefore, iC— — ~— 

A 

But W + iC 2 =CO\ 

That is, x" + Vt^l=(R- x )\ 

Or, x*=:%(R 2 -2Rx+x' i ). 

x'=SR 2 -6Rx. 

x=R(2V3-3). Ans. 



PROBLEM XIX 



In a right angled triangle, having given the perimeter, and 
the perpendicular let fall from the right angle on the hypo- 
tenuse, to determine the triangle; that is, its sides. 



237] 



GEOMETRY 



29 



Let ABC be the A , and represent its 
prime ter by p. Put AD— a, AB—x, 
AC=y. Then BC=p-x-y. 
G Because BAG is a right angle, 
oi?+y>=p*-2jp(x + y) + x* + 2xy + y' (1) 

BCAD=2 times the area of the triangle ABC. 
Also, AC'AB=2 times the area of the triangle ABC. 




Therefore, a(p—x—y)=xy. 

Keducing (1), 2p (x+y)=p* + 2xy 

Double (2), 2ap—2a(x-\-y)=2xy 

By subtraction, (2a + 2p)(x + y) — 2ap =p* 



Whence, 



x+y. 



p* + 2ap 

' 2a + 2p 



(2) 
(3) 
(4) 
(5) 

(6) 



Because BC=p—x—y, BC=p 



p*-\-2ap_ p' 



2a + 2p 2a+2p 



ap 1 



From (2) we observe that xy— 

2tO/-\- Z~p 

Equations (6) and (7), will readily give x and y. 



(7) 



PROBLEM XX. 

To determine a right angled triangle, having given the 
hypotenuse, and the difference of tivo lines drawn from the 
tivo acute angles to the centre of the inscribed circle. 

Let ABC be the tri- 
angle, and the center 
of its inscribed circle, 
AO and CO being joined, 
the triangle AOC is 
formed, and AO being 
produced to meet a per- 
pendicular from C, COD is an exterior angle. 

AO bisects the angle at A, and CO bisects the angle ACB. 




30 



KEY TO 



[2S7 



Therefore, the angle COD, is equal to half the angles A and 
C of the right angled triangle ABC. Consequently COD is 
45° ;D being 90°, the angle CD= 45°, whence OD=CD. 
Let CO=x, and AO — d+x, d being the given difference be- 
tween AO, OC. Also, let OD=y, and AC=h. 
Then 2y 2 =x* (1) 

Aff+DCf^^Iff, 
That is, ( x +y + dy + tf = 7i* (2) 

Expanding x*+y' z + d 2 + 2xy+2dx+2dy+y' i =h\ 
Substituting the value of 2y% as found in (1), and 
2x" + 2xy + 2d(x + y) = ¥ - d 2 

2x>+??i+2d(x+^)=h* 

V2 \ V2/ 

(2-f V2)x*+ (2 -fV2)a 7 x-A 5 -^ 



£ U + cfa: 



'2+Y2 



m. 



Whence, 






PROBLEM XXI 



To determine a triangle, having given the base, the per- 
pendicular and the difference of the tivo sides. 

Let ABC be the A. Put BD=x, 
DC=y, AC=z, AB=z+d, AD=a, 
BC—b. By the conditions, 

x + y=b (1) 

x 2 + a 2 =s 2 + 2<&-fd 2 (2) 

tf + a*=z* (3) 

By subtracting, x 2 — f — 2dz + d* (4) 

Factoring, (x+y)(x— y) — d{2z + d) 

That is, K x -y) = <^( 2s + «0' 




23T] GEOMETRY. . 31 

From this we have the proportion, 

b : (2z+d) : : d : (x—y) 
This proportion is the following rule given in trigonom- 
etry, viz. : 

In any plane, triangle, as the base is to the sum of the 
sides, so is the difference of the sides to the difference of the 
segments of the base. 

We return to the solution. From (1) we have, 

x—b—y, whence x 2 —y 2 = b 2 —2by. 
From (3) z= Vy*+a\ These values put in (4) give 
b 2 -2by=2dVy 2 + a 2 + d 2 
(b 2 -d 2 )-2by=2dVtfTa~ 2 
Squaring, (b 2 - dj - 4b (b 2 - d 2 )y + 4b 2 y 2 = 4d 2 y 2 +4a 2 d 2 
Or, (b 2 -dj-4b(b 2 -d 2 )y + 4(b 2 -d 2 )y 2 =4a 2 d 2 

{b>-tf)-4by+4y*=^ 

V_4by + 4y 2 = 1 ^ 2 + d 2 =m* 

Whence, b—2y=±m 

And, y=b+m, or b— m. 



PROBLEM XXII. 

To determine a triangle, having given the base, the per- 
pendicular, and the rectangle, or product of the ttvo sides. 

Let ABC be the a. Put BB=x, 
JDC—y, BC=b, AD=a, and the rec- 
tangle, (AB){AC)=c. 

Now in the right angled triangles, 
ABB, ABC, we have, 
A B = Vx 2 + a 2 . A C= Vf + a 2 . 




32 - key to [227 

Whence, (Vx 2 + a%Vj/ + a')=c (1) 

And, x-\-y—b (2) 

From(l), xhf + aXx'+y^+a^c' (3) 

From (2), x*+tf=V-2xy (4) 

This value substituted in (3), gives 
xy + a'b i -2a"xy-ha 4 = c i 

xY-2a'xy + a i = c 2 -a i b 2 

X y—a l =±Vc i —a 2 b' i 



xy=a*±Vc*-a*b* (5) 
Now equations (2) and (5) will give the values of x and y. 



PROBLEM XXIII. 

To determine a triangle, having given the lengths of the 
three lines drawn from the three angles to the middle of the 
opposite sides. 

Let ABC be the A. Place 
AD=a, BF=b, and OG=c. 

Also, put BD—x, one half of 
the base, BG—y, one half of the 
side AB, and FC=z, one half of ~~ ~~ D 

the side AG. Now by (Th. XLIL, B. I.), we have, 
ZB a +Ztf 2 =2#D 2 +2ZD a . 
That is Aif + 4z* = 2x* + 2a 2 (1) 

Similarly, 4x 2 + 4y*=2z' 2 + 2V (2) 

And, 4x 2 + Az 1 = 2tf + 2c a (3) 

Keduced sum, 6x 2 + 6f + 6s 2 =2(a* + V + c 1 ) 
Or, 2a; 2 + 2if + 2s 2 = f (a 2 + 6 2 + c 2 ) = ra 

From (1) 2?/ 2 + 22 2 =^ 2 + a 2 

By subtraction 2x' i =— x* + m— a* 




Or, 2>x^—m— a\ Whence, x-- 



Vs 



«J37] 



GEOMETRY. 



33 



In like manner we find y- 



Vm—{ 



V3 



And, «= — . Double of these values will be the 



V3 



sides required. 



PROBLEM XXI Y. 




In a triangle having given the three sides, to find the 
radius of the inscribed circle. 

Let ABO be the A. A 

From the center of the 
circle 0, let fall the per- 
pendiculars OG, OF, OB, 
on the sides. 

These perpendiculars 
are all equal, and each 
equal to the radius re- 
quired. 

Let the side opposite to the angle A, be represented by a, 
the side opposite B, by b, and opposite O, by c. Put OF, 
OB, etc., equal to r. 

It is obvious that the double area of the A BOO is ex- 
pressed by ar ; the double area of A OB by cr ; the double 
area of AOG by br. Therefore, the double area of ABO is 

(a+b + c)r. 
From A let drop a perpendicular on BO, and call it x. 
Then ax= the double area of ABO. Consequently, 

(a + b-\-c)r=ax (1) 

The perpendicular from A will divide the base BO into two 

3 



34 



KEY TO 



[237 



segments, one of which is Vc*— x% the other, \ V— x\ and 
the sum of these is a ; therefore, 

V^^+VV-^^a (2) 

Vc*—x 2 =a— Vb'—x* 

c i -x i =a' 2 -2aV¥^c?-\-b' i ~x i 

2aVb T ^x*=d i + V-c' i 



Vb 2 - 



x 



a' + V-c 



2a 



-=m 



Whence, b' l —x 2 —m % 

Or, x=V¥- 

This value of x put in (1), gives 



•m 



(a + b + c)r=aVb' i — 



m 



Whence, 
the required result. 



r—- 



aVtf-m* 



a+b+c ' 



PROBLEM SIT, 



To determine a right angled triangle, having given the 
side of the inscribed square, and the radius of the inscribed 
circle. 



Let AB be one side of the triangle, 
and BG a side of the inscribed square, 
which we designate by a. And let 
BH=r, a radius of the inscribed 
circle. 

Then GH=a-r=d. 

Join AE, touching the circle at K, 
and BD, touching the circle at P. 




2B7] GEOMETRY. 35 

Conceive the lines AE and BJD produced, meeting at a 
point G, then ABG is the triangle sought. 
Let AH=AK=x, and KG=GP=y. 

Then AG— the hypotenuse=0 + ?/). (Prob. XV.) 

AB=x + r, and GB—y + r. 

(x + ry + (y + ry = tx + yy (1) 

Again AG=x— d, GE=a, and GB—y—d. 

Now, AG : GE : : ED : GJD 

That is, (x—d) : a : : a : y—d. 

Whence, xy—d(x + y) + d 2 =a i (2) 

Expanding and reducing (1), we obtain, 

2r(x + y) + 2?' 2 =2xy 
Or, xy=r(x + y) + r* (3) 

From (2) xy—d(x + y)+a 2 —d' 2 

By subtraction = (r — d) (x -f ?/) + r 2 + cf — a 2 

r*4-d 2 —a* 
Whence, x-\-y— — . (5) 

This value of (x-\-y) placed in equation (3) and reduced, 
and we obtain, 

r(d 2 — a? + dr) , cs 

Equations (5) and (6) will give separate values of x and y, 
and thus solve the problem. 



PROBLEM XXVI. 

To determine a triangle and the radius of the inscribed 
circle, having given the lengths of three lines drawn from 
the three angles to the center of that circle. 



36 



KEY TO 



[5137 




Let ABC be the A, the cen- 
ter of the circle. 

Put AO=a, OB=b, OC=c. 
AO bisects the angle A. 

Produce AO to D. Then be- 
cause the angle A is bisected, 
D nx - CD : DB : : AG : AB. 

Put AB=x, AC=y, and let the ratio of AB to ^i) be 
n ; then nx=BD and ny—CD. 

Now as the angle (7 is bisected by CO, we have, 

AC : CD : : ^40 : OB 
That is, y : ny : : a : OB 

Whence, OD = na. 

Because AB bisects the angle A, we have, (Th. XX. 
Book III.) 

xy = a 2 (1 + iif + tfxy 



Also, 
And, 

From (1), 



nx 2 =b 2 + na* 



ny 1 -. 
xy-. 



■ c +na 

a\l+rCf a\l + n) 



(1) 
(2) 
(3) 



1 



■n 



(4) 

(5) 



1-n 2 
The product of (2) and (3), gives 

n 2 x 2 y 2 =:{c 2 + na 2 )(b 2 + na 2 ) 
Squaring (4), and multiplying the result by n*, also gives 

(6) 

V x — ,oj 

Equating (5) and (6), gives 

(c 2 + na 2 )(b 2 + na 2 )(l-n) 2 =aXl + n 2 )n 2 

This equation contains only one unknown quantity n } 
but it rises to the fourth power — hence this problem is not 
susceptible of a solution under this notation, short of an 
equation of the 4th degree. 






&a7] 



GEOMETRY, 



37 



When a, b, and c, are numerically given, eases may occur 
in which the resulting equation may be of a low degree. 
When b—c, then x—y. 

The three sides being determined, the radius of the in- 
scribed circle is then found by Problem XXIV. 



PROBLEM XXVII. 



To determine a right angled triangle, having given the 
hypotenuse, and the radius of the inscribed circle. 

Let ABC be the A. Place A 

AE=x, EB=EO=r, and CF=y. 
Then, AB=x+r 

BC=y+r 

And, AC=(x + y)=h. 

Now, (x + ry + (y + ry=(x + y)\ 

Expanding and reducing the above, we have 

r(x + y)+r*=xy 
That is, xy=r* + hr 

And, x + y=h. 




Whence, 
And, 



x=lh + iVh*—4hr—4r 



y=ih-Wh*-4/ir-4r* 



PROBLEM XXVIII. 

Here the problem is given in general terms— this same 
problem is numerically given in (Book V. Prob. 8), and is 
solved in this key, on page (7). 



38 



KEY TO 



[237, 838 



PROBLEM XXIX. 



This problem involves the same relations as (Prob. 12, Book 
V.), which is solved in this key, on page (9). 



PROBLEM XXX. 

The radius of a circle being given, also the rectangle of 
the segments of a chord, to determine the distance from the 
center to the point at ivhich the chord is divided. 

„-.--- -_ % Let C be the center of the cir- 

/' \ cle, AB the chord, and P the given 

point in it. Join CP and denote 
that distance by x. 

Through P draw the chord DH 
at right angles to PC. Then be- 
cause CP is a line from the cen- 
ter perpendicular to the chord 
DH, it must bisect that chord. (Th. I. Book III.) 
DPCis a right angled triangle, JDC=r, then 




DP=Vr 2 — x\ 

But because DH, and AB intersect at P, we have, 
AP.PB=DP.PH=DP* 

Whence, AP.PB=r*—x* 

Or, x=Vr r -AP.PB. 

That is the distance from the center to the required point 
will be found by subtracting the rectangle from the square 
of the radius of the circle, and extracting the square root of 
the remainder. 



^38] 



GEOMETRY. 



39 



PROBLEM XXXI. 

If each of the tivo equal sides of an isosceles triangle he 
represented by (a), and the base by (2b), what will be the 
radius of the inscribed circle in terms of a and b. 

a Let ABC be the trian- 

gle, and 
AB=AC=a< BD=b. 




ThenAD=Va'-b\ 
Place OD=B=HO. 
Now, by the similar 
triangle A HO, ABB, we 



That is, 
And, 

Whence, 



bVa 
B= 



have 
c AOiOH 

AD-B : B 
-bB = aB 



\ABiBB 

: a : b 



bVa 2 -b n - bVa—b 



a+b Va+b 



PROBLEM XXXII. 

From a point without a circle whose diameter is (d), a 
line equal to (d) is drawn to the concave terminating in the 
concave arc, and bisected at the first point of meeting the 
circumference. Bequired the distance of the point without 
from the center of the circle, 

y ''*' ~"^*x Let G be the center of the 

/ \ circle, and B a point with- 

* out, and AP—d, the diam- 

* eter of the circle, which is 
bisected by the circurnfer- 

/' ence at B. 

Join CA, CB, and OP. 
Because AP is equal to the diameter of the circle, and it is 




40 KEY TO [238, g40 

bisected in B, AB must be equal to the radius of the circle ; 
that is, AB—BG=GA, and the triangle ABC equiangular. 

Also BP=BG, therefore the angle BGP=BPG. But 
the angle CBP = 120°, therefore the angle at P is- 30°, and 
the angle PGA =90°. 

Hence, GA*+Pff=PA\ 

Or, (±dy + CP 2 =d\ 

Whence, GP' = \d\ or, GP=iVl. 

the value of the line sought. 



MISCELLANEOUS PROPOSITIONS. 
(Page 238.) 

(3.) If from any point ivithout a circle, two straight lines 
be drawn to the concave part of the circumference, making 
equal angles with the lines joining the same point and the 
center ; the parts of these lines which are intercepted loilhin 
the circle, are equal. 
p Let P be any point without a circle, 

and C the center of the circle. Join 
PG and draw PA, PB making equal 
angles with PG. 

Join GA, GB, Ga and Gb. We 
are to prove that Aa=Bb. 

In the two triangles PGA, PGB, 

we have GA = GB, and PG common 

A " to both triangles, and the angle GPA 

opposite the side AG is equal to GPB opposite the equal 

side BG. 

That is, the two triangles have three parts respectively 




240] GEOMETRY. 41 

equal. Therefore, the three other parts are also equal, and 
PA=PB. In like manner by the two triangles PGa, PCb, 
we prove Pa=Pb. 

Whence, by subtraction, 

PA-Pa=PB-Pb. 

That is, Aa=Bb, and the theorem is demonstrated. 

(4.) If a circle be described on the radius of another 
circle, any straight line drawn from the 'point where they 
meet, to the outer circumference, is bisected by the interior 
one. 

Let AG be the radius 
of one circle and the di- 
ameter of another, as 
represented in the figure. 
From the point of con- 
tact A, of the two cir- - c 
cles, draw any line, as AH ; this line is bisected in D. 
Join DG and HB. Then ADG being in a semicircle, is a 
right angle ; also, AHB is a right angle, for the same reason : 
therefore, DG and HB are parallel. Whence, 
AD : AH : : AG : AB 

But as AB is the double of AG, therefore AH is the double 
of AD, or AH is bisected in D, which was to be shown. 

(5.) From two given points on the same side of a line 
given in position, to draw two straight lines, which shall con- 
tain a given angle and be terminated in that line. 

Let AB be the line given in position, and P and p the 
given points. To make the problem definite, we take the 
given angle at 35°. 




42 



KEY TO 



[240 



Join Pp, and draw PpQ a right angle. Take 35° from 

90°, and the remainder is 55°. 
Now, at P, make the angle 
pPQ equal to 55°, then the 
third angle of the triangle 
PQp will be 35°, the given 
angle. But it is not located 
in the line AB. 

About the triangle PQp, de- 
scribe the circle, and whence it 
cuts the lines AB, draw PA, pA, and the angle PAp=PQp, 
because they are angles in the same segment of a circle ; and 
the angle is located as was required. The point B is another 
in the line when the angle would be the same in magnitude. 

Scholium. — The given angle must be less than the angle 
PCp, otherwise the two lines from P, p, making the angle, 
would meet before they could reach the line given in position. 




(6.) If from any point tvithout a circle, lines be draivn 
touching it, the angle contained by the tangents is double 
the angle contained by the line joining the points of contact, 
and the diameter draivn through one of them. 

p 

Let P be a point without a cir- 
cle, and PA, PB, tangents to the 
circle. Lines joining the center 
to the point of tangency, make 
right angles with the tangent lines. 
(Th. IV., Book III.). Join also 
PC and AB, PC is common to 
the two right angled triangles PA 
andP^C. Hence, PB=PA, and 
the angle at P is bisected by PC. 




£40] 



GEOMETRY 



43 



AB is a line drawn from the right angle of the triangle 
PA G perpendicular to its hypotenuse PG, and divides that 
triangle into two similar triangles (Th. XXV., Book II., 
1). Hence the angle BAG is equal to the angle APG, or 
BPA, is double of the angle BAG, which was to be demon- 
strated. 



(7.) If from any two points in the circumference of a cir- 
cle, there be drawn two straight lines to a point in a tangent 
to that circle, they will make the greatest angle when drawn 
to the point of contact. 

Let A and B be the 
two points in the circle, 
and GD a tangent line. 
The proposition requires 
us to demonstrate that 
the angle AGB is greater 
than the angle ABB. AGB=AOB, (Th. IX., Book III., 
Cor.) But AOB is greater than ABB, (Book I., Th. XII., 
Cor. 1). Therefore, AGB is also greater than ABB. 




(8.) From a given point within a given circle, to draw a 
straight line which shall make with the circumference an 
angle less than any angle made by any other line drawn 
from that point. 

Note. — An angle between a chord and a circumference is the same as be- 
tween the chord and a tangent drawn through the same point. Thus the 
angle made by the chord AB, and the circumference at B is the same as be- 
tween AB and a tangent drawn through B. 



Let G be the center of the circle, P a given point within 



44 



KEY TO 



[240 




the circle not at the center. Join CP, and through P at 
right angles to CP draw the chord AB. 

The angle made between AB and 
the circumference B is measured by 
by half the arc BGA, and this is 
less than any other angle made by 
any other line drawn through P, 
and the circumference. 
Through P draw any other line as FPG. Now the angle 
which PG makes with the circumference is measured by half 
the arc GAF. But GAF is a greater arc than BGA, because 
GF is greater than AB. GF we know to be greater than 
AB, because CD is less than CP, CP being the hypotenuse 
of the right angled triangle CDP. Consequently the an- 
gle which PB makes with the circumference is less than 
that which PG makes with the circumference. 




(9.) If two circles cut each other, the greatest line that can 
be drawn through either point of intersection, is that ivhich 
is parallel to the line joining their centers. 

>X Let c and C be 

the centers of two 
circles which cut 
each other, and 
A, one of the points of intersection. Now our object is to 
prove that a line drawn through A, parallel to Cc, will be 
greater than any other line, as GB drawn through A, and 
not parallel to cC. 

From cC draw perpendiculars ccl, CD, to GB. AB is 
bisected in D, and AG in d, therefore, dD, is half GB. But 
dD is less than dm=cC, because dm is the hypotenuse of 
the right angled triangle dDm. 




240] GEOMETRY. 45 

But if GB were drawn parallel to cO through A, then 
dD would be equal to cC. Consequently a line drawn 
through A, parallel to cC, is the greatest possible through 
A, which was to be shown. 



(10.) If from any point within an equilateral triangle 
'perpendiculars be drawn to the sides, they are together equal 
to a perpendicular drawn from any of the angles to the op- 
posite side. 

Let ABG be the equilateral A, CD 

a perpendicular from one of the an- 
gles on the opposite side ; then the 
area of the A is expressed by \AB 
x CD. Let P be any point within 
the triangle, and from it let drop the 
three perpendiculars PG, PR, PO. G D 

The area of the triangle APB is expressed by \AB x PG. 
The area of the A CPB is expressed by \CBxP0: and 
the area of the A CPA is expressed by \CA x PH. By 
adding these three expressions together, (observing that CB 
and CA are each equal to AB,) we have for the area of the 
whole A ACB, \AB(PG+PH+P0.) 

Therefore, \AB x CD=\AB(PG+PH+PO.) 

Dividing by \AB, gives CD=PG+PH+P0. 

(11.) If the points bisecting the sides of any triangle be 
joined, the triangle so formed will be one-fourth of the given 
triangle. 

If the points of bisection be joined, the triangle so formed 
will be similar to the given A, (Th. XX., Book II.) 

Then, the area of the given a will be to the area of the 
A formed by joining the bisecting points, as the square of a 



46 



KEY TO 



[240, 241 



line is to the square of its half ; that is, 2 2 to 1, or as 4 to 
1. Hence the a formed is J of the given triangle. 

(12.) The difference of the angles at the base of any tri- 
angle, is double the angle contained by a line drawn from 
the vertex perpendicular to the base, and another bisecting 
the angle at the vertex. 

Let ABC be a a . Draw AM 

bisecting the vertical angle, and 

draw AD perpendicular to the 

base. 

The theorem requires us to 

prove that the difference between the angles B and C is double 

of the angle MAD. 

By hypothesis, the angle CAM—MAB. 

That is, CAM=MAD + DAB. (1) 

_ ■ ^ TT A tC+CAM+MAD=WP\ (2) 

By ( Tl,Xn.,B.L,Co,4)^ + ^ = ^ ^ 

Therefore, B+DAB=C+ CAM+ MAD (4) 

Taking the value of CAM from (1), and placing it in (4), 
gives B+DAB=C+MAD+DAB + MAD. 

Keducing (B—C)—2MAD, which verifies the text. 




(13.) If from the three angles of a triangle, lines be drawn 
to the middle of the opposite sides, these lines will intersect 
each other in the same point. 

Let ABC be a A , bisect 
BC in F, AC in F. 

Join AF and BF, and 
through their point of in- 
tersection 0, draw the line 
CD. Noiv if toe prove AD=DB, the theorem is true. 




241] GEOMETRY. 47 

Triangles whose bases are in the same line, and vertices in 
the same point, are to one another as their bases ; and when 
the bases are equal, the triangles are equal. For this reason 
the aAFO=aFCO, and the aCOE= aEOB. 

Put AAFO=a ; then AFCO=a. Also, put ACOE=b, 
as represented in the figure. 

Because CB is bisected in F, the A ACE is half of the 
whole A ABC. Because AC is bisected in F, the A BFC is 
half the whole A ABC 

That is, 2a + b=2b + a. 

Whence, a=b, and the four triangles above 

the point are equal to each other. 

Let the area of the A ABO be represented by x, and the 
area of DOB by y. 

Now taking CD as the base of the triangles, we have 





2a : x : : CO 


OD 


Also, 2b = 


2a : y : : CO 


OD 


Whence, 


2a : x : : 2a : 


y. Or, x 


Therefore, 


AD=DB. 





Scholium. — If the triangle, ABC be regarded as a thin 
lamina of matter, its center of gravity must be somewhere 
in the line AE ; for the part AEC~ the part AEB. For 
a similar reason, the center of gravity must be somewhere in 
the line BF. Hence it must be at their intersection, 0. 

Again, the triangle AEB— the triangle ADC And since 
x=y } and a =b, we have, 

2x + a=2a+x, or, a—x. 

Hence, aAOB=\aACB ; and OD=±CD. 

(14.) The three straight lines which bisect the three angles 
of a triangle, meet in the same point. 



48 



KEY TO 



[241 




Let ABC be any 
triangle, "bisect two of 
its angles A and B, the 
bisecting lines meeting 
at 0. 

Let fall perpendicu- 
lars from on the 
three sides, OD, OG, OH. 

Join CO, and now if we can demonstrate that CO bisects the 
angle C the proposition will be proved. The two right angled 
triangles AOD, AOG have a common hypotenuse, AO, and 
equal angles respectively, therefore, OD=OG. In like man- 
ner, by the two right angled triangles, DOB and OHB, we 
prove OD=OH. Whence, GO— OH, and the hypotenuse CO 
is common to each of the triangles, CGO and CHO, therefore 
their angles are equal, and the angle C is bisected by the line 
CO. Hence the bisecting lines all meet at the point 0, and 
the proposition is proved. 

Cor. — Let the student observe that OD, OG, OH, each, is 
equal to the radius of the inscribed circle. 



(16.) The figure formed by joining the points of bisection 
of the sides of a trapezium, is a parallelogram. 

Let ABCD be a trapezium. Draw 
the diagonals AC, BD. Bisect the 
sides in a, b, c, and d. Join abed. 
We are to prove that this figure is a 
parallelogram. 

ABD is a a whose sides are bisected in a and b ; there- 
fore, ab is parallel to BD, (Th. XVII., Book II.) In the 
same manner we can prove that do is parallel to BD. Con- 
sequently ab and dc are parallel. It may likewise be shown 




241] 



GEOMETRY. 



49 



that ad and be are parallel. Hence the figure abed is a 
parallelogram. 



(17.) If squares be described on the three sides of a right 
angled triangle, and the extremities of the adjacent sides be 
joined, the triangles so formed are equivalent to the given 
triangle, and to each other. 

Let^^Cbethe 
original right an- 
gled triangle ; on 
its sides describe 
the squares AD, 
BG, and AH. 

Join KE, DF, 
HG, thus form- 
ing the triangles 
KAE, DBF, and 
HOG, and we are 
to prove each of them equal to the triangle ABO, and to 
each other. 

We will now prove the triangle ALK equal to ABC, as 
follows, 

The angle LAB=90°, 

Also KAC=90°. 

From each of these equals take away the common angle 
LAC, and the remaining angles KAL and CAB must be 
equal. Again, ALK is right angled at L, and ABC right 
angled at B, and the hypotenuse AK of the triangle ALK, 
is equal to AC the hypotenuse of ABC Therefore, ALK 
is equal to ABC 

Whence, AL=AB=AE, and KL=BC 

Now, because AL—AE, the area of the triangle KAE is 




50 KEY TO [241 

equal to the area ALK, or ABC, and this is one of the facts 
which was to be demonstrated. 

In like manner because A CH =90°, and BCM=90°, taking 
away the common angle MCA from each of these equals, we 
have the angle HCM=ACB; but HC= AG and the trian- 
gles right angled, one at 31, the other at B. Therefore, the 
triangle HCM is equal in all respects to ABC. 

That is, CM=CB, but CB=CG, therefore MC= CG; con- 
sequently the area of the triangle HCG— the area HCM, 
or ABC 

Now, because the area KAL—ABC 

And the area HCG=ABC 

Therefore, area KAL= the area HCG. 

The triangle BBF—ABC, because the opposite angles at 
B are equal, each 90°. AB=BD, and CB^BF ; and the 
proposition is fully demonstrated. 



(18.) If squares be described on the hypotenuse and sides 
of a right angled triangle, and the extremities of the sides 
of the former, and the adjacent sides of the others, be joined, 
the sum of the squares of the lines joining them, ivill be five 
times the square of the hypotenuse. 

The two trian- 
gles KLE, and 
HMG are right 
angled triangles, 















L 




A. 


H/ 




M / 


B 






C 








^^^ 










^^ 



right 



angles at L, 



and at M. Ob- 
serve also that 
AH=AB=AL. 
Whence, 
HL=2AB. 



341] ■ GEOMETRY. 51 

Also, KL=BC, BC=CG=C3L Whence, MG=2BC. 

But W+KL 2 =EIL* 

And MG*+WH 2 =£G* 

That is, (2ABy + (BC'y=(FKy (1) 

And (2£Q 2 + (^£)^(Zr£) 2 (2) 

By adding (1) and (2), observing that (2ABy is 4AB 2 

5AB 2 + 5BC 2 = (£7Q 2 + (HG)\ 
That is, 5 times (AG)'=(EK)*+(HG) % 
which verifies the proposition. 

Cor. — The area of the whole figure is double the square 
on AC, and 4 times the area of the triangle ABC. 

(19.) The vertical angle of an oblique angled triangle, in- 
scribed in a circle, is greater or less than a right angle, by 
the angle contained between the base and the diameter drawn 
from the extremity of the base. 

Let ACB be the vertical 
angle of a triangle, its base 
the chord AB. From the ex- 
tremity of the base B, draw 
I the diameter of the circle BD 
Join CD. 

Because BD is a diameter 
the angle BCD is a right an- 
gle ; therefore, the angle ACB 
is greater than a right angle by the angle A CD. But A CD 
=ABD, each being measured by half the arc AD. 

That is, ACB is greater than a right angle by the angle 
ABD, which was to be demonstrated. 

Or, let A'BC be the triangle in a circle, and A'CB its ver- 
tical angle. This is less than a right angle, by the angle 
A ] CD, or A'BD. 




52 



KEY TO 



[2M 




(20.) If the base of any triangle be bisected by the diame- 
ter of its circumscribing circle, and from the extremity of 
that diameter, a perpendicular be let fall upon the longer 
side, it ivill divide that side into segments, one' of ivhich 
ivill be equal to half the sum, and the other to half the differ- 
ence of its sides. 

Let ABO be the A ; bisect its 
base by the diameter of the cir- 
cle drawn at right angles to AB. 
From the center let fall Om 
at right angles to AC, it will then 
bisect AC. From the extremity 
of the diameter H, draw Hh per- 
pendicular to AC, and conse- 
quently parallel to Om. Produce Hh to M and join ML. 
Complete and letter the figure as represented. 

The two triangles Aab and Hha are equiangular. The 
angle a is common to them, and each has a right angle by 
construction, therefore the angle H— the angle A, and the 
arc ML = the arc CB, (Th. IX. ? Book III., C.) ; therefore 
OB— ML. The angle HML is a right angle, because it is in 
a semicircle, therefore, ML is parallel to AC, and ML is bi- 
sected in n. 

Now Am=±AC nL=md=iML=±BC. 
Therefore by addition, Am + md=\(AG+CB). 
Or, Ad=i(AC+CB.) 

But, if Ad is the half sum of the sides, dO or Ah must 
be the half difference ; for the half sum and half difference 
make the greater of any two quantities. 

(21.) A straight line drawn from the vertex of an equi- 
lateral triangle, inscribed in a circle, to any point in the 
opposite circumference, is equal to the sum of the two lines 
drawn from the extremities of the base to the same point. 



£41] 



GEOMETRY 



53 



Let ABC be an equilateral triangle inscribed in a circle, 
and designate each side by a, 
as represented in the figure. 

Take any point D on the arc 
BC, and join AD and designate 
it by x. Join BD and DC. 

We are required to demon- 
strate that x—BDAtDC. 

Observe that A BDC is a quad- 
rilateral inscribed in a circle, and 
by (Th. XXI., Book III.), we 
must have, 

ax=a(BD)+a(DC). 

Dividing each side by a, and x=BD-\-DC, which was to 
be demonstrated. 




(22.) The straight line bisecting any angle of a triangle 
inscribed in a circle, cuts the circumference in a point equi- 
distant from the extremities of the side opposite to the bi- 
sected angle, and from the center of the circle inscribed in 
the triangle. 

Let ABC be a triangle in 
a circle, and draw CD bi- 
secting the angle at C, and 
draw BG bisecting the an- 
gle at B. The intersection 
of these two bisecting lines 
at 0, is the center of an 
imaginary circle inscribed 
in the triangle ABC. 

The line CD bisects the 
angle at C by hypothesis, therefore, the arc AD=z the arc 




54 



KEY TO 



[«4fc 



BD. Hence the chord AD= the chord BD, and thus the 
first part of the theorem is proved. 

In like manner, because BG bisects the angle B, the arc 
AG= the arc GO. 

Now in the triangle DBO, the angle DBO or DBG is 
measured by half the sum of the arc, AD -{-AG. And the 
angle DOB is measured by half of the sum of the arcs 
DB+GC But AD + AG=DB+GC. Therefore, the angles 
DBO and DOB are equal, and the triangle is isosceles, and 
DB=DO. 

In like manner we can prove that DA— DO, and the 
second point of the theorem is demonstrated. 



(23.) If from the center of a circle, a line be drawn to any 
"point in a chord of an arc, the square of that line, together 
with the rectangle of the segments of the chord, ivill be equal 
to the square on the radius of the circle. 

Let O be the center of a 
circle, and P any point in 
the chord AB. Join PC, 
then we are required to 
prove that 

PC n + AP.PB=CQ\ 
Through P draw the chord 
RQ at right angles to PC, 
and join CQ. 

Because CP is a line perpendicular from the center to a 
chord it bisects the chord. Hence, BP—PQ. 
And because two chords intersect at P, we have ; 
AP.PB=BP.PQ=~PQ\ 




242] GEOMETRY. 55 

Add CP* to each member of this last equation, and we 
have, 

CP* + AP.PB=PQ> +P(f 

But by the right angled triangle PCQ, we perceive that the 
second member is equal to CQ\ Therefore, 

CP°' + AP.PB=:CQ\ 

which was to be demonstrated. 



(24.) If two points be taken in the diameter of a circle 
equidistant from the center, the sum of the squares of the 
tivo lines drawn from these points to any point in the cir- 
cumference, will always be the same. 

Take the same circle as in the preceding theorem, and let 
GC— CH. Take any point in the circumference as D, and 
join DG, DC, DH. 

Now because the base of the triangle GDH is bisected in 
C, we have by (Th. XLIL, Book I.), 

GD*+DH*=2DC' i +2GC\ 

But at whatever point on the circumference D may be placed, 
DC and GC will always retain the same value. Therefore, 
the sum of the squares of the other two sides is constantly 
equal to the same sum. 

(25.) If on the diameter of a semicircle, two equal circles 
be described, and in the space enclosed by the three circum- 
ferences, a circle be inscribed, its diameter will be two thirds 
the diameter of either of the equal circles. 

Let AB be the diameter of one of the two equal circles 
and construct the figure as here represented. 



5Q 





** 


« r-^" ■*"" " 


, 






,*" 


/ 




X 








J 




t » 


1 




/ 




1 
1 

\ 


/ 


/ «.«-"" " ' 






/ *■* 


" "* * »\ 




/ X* 




x / 


/,' 




/v._ 


// 




\ 


ft 




\ 


1 

\ 

1 > 




\ 
\ 
1 

t 



KEY TO 
E 



[242 



We are required to show 
the relation between CB and 
ED. 

Place CB =r and ^Z>=a\ 

Then BE=2r, and £Z>= 
2r— a, CD=r-lx. 

In the right angled trian- 
gle CBD, we have, 

cb 2 +bd*=cd\ 



A. C B 

That is, r 2 + (2r-a) 2 = (V + a) 2 , 

Or, r 2 + 4 r 2 _4 ra; + ^ =r 2 + 2ra; + ^ 2 

Whence, 6nc=4r 2 , and a5=fr 



(26.) 2jf a perpendicular be drawn from the vertical an- 
gle of any triangle to the base, the difference of the squares 
of the sides is equal to the difference cf the squares of the 
segments of the base. 

Let ABG be a triangle. From 
A let fall AD perpendicular on BC, 
making two right angled triangles 
ADB, ADC. These triangles give 

(ADy+(DBy=:(ABy 
(ADy+(pcy=(Acy 

By subtraction (BD) 2 -(DOy = (ABy-(ACy. 
This equation demonstrates the theorem. 

Cob. — By factoring the above equation, we have, 

(BD+DC)(BD-DC) = (AB+AC)(AB-AC). 
Observing that BD^-DC—BC, and changing this equation 
into a proportion, we have, 

BC : AB+AC=AB-AG : (BD-DC). 




&4&] GEOMETRY. 57 

(This is Prop. 6, Plane Trigonometry, Page 256, Geo- 
metry.) 

This proportion is true whatever relation exist between 
AB and AG. When AB=AC, then BB = DO and the 
preceding proportion becomes 

BO : 2AB : : : 0. 

This is an apparent absurdity, but we can reconcile it to 
truth by taking the product of the extremes and means, and 
we have, 

(BG)0=<(2AB)0. 

An equation obviously true, 

BC 

Or, — -, and divided by is any quantity whatever, 

Hence, : : : a : to any quantity whatever. 



(27.) The square described on a side of an equilateral 
triangle is equal to three times the square of the radius of 
the circumscribing circle. 

Let ABC be an equilateral trian- 
gle. Let fall the perpendicular 
AE on the base, that line will bi- 
sect the angle A. Draw BD bi- 
secting the angle at B. 

We will now prove AB—BB ; 
then D must be the center of the 
circumscribing circle. 

Each angle of an equilateral triangle is 60° (J of 180°), 
Bisecting each of these angles, we have, BBA, BAB, each 
30°, and therefore AB—DB, and hence, if D be taken as 
the center of circle with DB or BA as radius, that circle 
will circumscribe the triangle. Put AB=2a, then BE=a. 




58 



KEY TO 



[242 



Also, place BB—x, then BE—\x, (It being the side of a 
right angled triangle opposite 30°). (Prob. 2 of this Key.) 
Now by the right angled triangle BBE, we have, 
(BEy + (BE)*=(BB)\ 



That 



is, 



a i -\-lx 2 =x% or, 3x i =4a 9 . 



But (4a*)=(ABy 
strates the theorem. 



That is, 3x* — {AB)\ which demon- 




(28.) The sum of the sides of an isosceles triangle, is less 
than the sum of the sides of any other triangle on the same 
base, and between the same parallels. 

tj Let ABO be the isosceles trian- 

gle. AB=AC Through the point 
A draw GAR parallel to BC. 

Take G any other point on the 
line GH, and draw BG and GC. 

We are to show that AB+AO 
is less than BG+GC. Produce 
AB to B, making AB=AB, or AG 
Then by reason of the parallels GH and BC, the angle 
D AH is equal to the angle ABC, and HAC=ABC 
Because AD=AC, the angle ADH= the angle ACH. 
Whence the two triangles ADH and ACH, are equal in 
all respects, and GH is perpendicular to DC ; whence any 
point in the line GH is equally distant from the two points 
D and C 

Now the straight line BB=BA + AC, and because JDG= 
GO, BG+GB=GB+GG But BG+GB, the two sides of 
a A are greater than the third side DB ; therefore GB-\-GC 
is greater than BB, that is, greater than BA +AC 



242] 



GEOMETRY, 



59 



(29.) In any triangle, given one angle, a side adjacent to 
the given angle, and the difference of the other tivo sides, to 
construct the triangle. 

Draw AB equal to the given C 

side. From one extremity of 
the base A, draw AC indefi- 
nitely, making the angle BAG, 
equal to the given angle. (Prob. 
V., Book IV.) 

Take AD equal to the given difference between the other 
two sides and join DB. From the point B make the angle 
DBC equal to the angle BDC, and the point C will be the 
vertex of the triangle required. Because BC= CD, and AD 
was made equal to the given difference of two sides, and AB 
the given side. 




(30.) In any triangle, given the base, the sum of the other 
two sides, and the angle opposite the base, to construct the 
triangle. 

Take any point A, and from it 
draw the line AD, equal to the 
sum of the sides. 

At the extremity D, make the 
angle ADB equal to half of the 
given angle opposite the base. 
From A as a center, with the dis- 
tance Am equal to the given base, describe an arc, mnB 
cutting DB in n and B. 

From B draw BC making the angle DBC, equal to the 




60 KEY TO [242 

angle D. Also from n draw nC ] parallel to BG, then ABG 
and AnC', are triangles corresponding . to the given condi- 
tions. 

For, AB=An=Am, the given base. The angles AGB, 
or AGhi opposite the respective bases, AB, An, are equal 
to the given angle, because A OB being the exterior angle of 
the triangle GDB, is equal to the sum of D and DBG ; but 
DBG was made equal to D by construction, and D was made 
equal to half the given angle. 

The same reasoning applies to the triangle DG'n. 

Also, GD—CB by construction, therefore AG-\-GB— 
AG+GD, or AD the sum of the given sides. 



(31.) In any triangle, given the base, the angle opposite 
to the base, and the difference of the other two sides, to con- 
struct the triangle. 

Let ABG be the triangle. 

To discover its proper or direct 

construction, we must suppose 

the problem solved, and then 

7 a analyze it. 

If AB is the given base, and AD the given difference of 

the other sides, it is then obvious that DG=GB, and knowing 

the angle G which is given, w r e can determine each of the 

other angles of the triangle GBD. 

Therefore, GDB is known, and its supplement ADB is 
known. Whence the angle 

ADB=90° + W. 
Take any point A and draw AG indefinitely. From the 




^4Q] GEOMETRY. 61 

scale of equal parts take AD = to the given difference, and 
make the angle 

ADB=90° + ±O. 
. Take the given base from the scale of equal parts in the 
dividers, and with one foot on A as a center, strike an arc, 
cutting DB in B. The line AB will be the base. 

From the point B, and from the line BD make the angle 
DBO=BDO, and produce BO and BO until they meet in 
0, and we have the triangle ABO, as was required. 



PLANE TRIGONOMETRY. 



SECTION II. 
(Page 289.) 

Note. — In each of the following examples in this section, there are six parts, 
three angles A, B, and C, and the three corresponding sides a, b, c, A being 
opposite to a, etc. Each of the following examples may be referred to one and 
the same triangle. The right angle always at B, and in right angled trigo- 
nometry this part is always given, and not generally expressed. The right 
angle and two other parts being given, the remaining three parts can be deter- 
mined. 

Ex. % Given AC, 73.26, and 
the angle A, 49° 12' 20", to find 
the other parts. 

From 90° take A, and we have 
0=40° 47' 40". 




As sin. 90°. 
To sin. 49° 12' 20" 
So is 73.26 
To BC, 55.462 



10.000000 
9.879129 

1.864867 
1.743996 



cos. 
los:. 



10.000000 
9.815144 

1.864867 
1.680011 



An& BO, 55.46. AB, 47.87. 



Ex. 3. Given AB, 469.34, and the angle A, 51° 26' 17", 
to find the other parts. 



^§9] PLANE TRIGONOMETRY. 63 

From 90° take A, and C=38° 33' 43". 

For BC. For AC. 

As sin. C, 38° 33' 43" 9.794739 sin. C 9.794739 

Sin. A 9.893171 sin. 90° 10.000000 

AB, 469.34 2.671488 AB 2.671488 



(Sin. A) {AB) 12.564659 AC 2.876749 

BC, 588.74 2.769920 AC 752.92. Ans. 

Ex. 4. Given AB, 493 ; and the angle C, 20° 14', to find 
the other parts. 

The remaining angle A is of course, 69° 46'. 

ForCB. For AC 

As sin. C, 20° 14' 9.538880 sin. C 9.538880 

Cos. C, or sin. A 9.972338 ) sin. 90 10.000000 

AB, 493 2.692847 ) AB 2.692847 



12.665185 AC, 1425.5=3.153967 
CB, 1337.53 3.126305 

Ex. 5. Given AB, 331, and the angle A, 49° 14', to find 



he other parts. 






The angle (7=40° 46'. 






For C%. 




For AC. 


As sin. C 9.814900 


sin. C 


9.814900 


Cos. C or Sin. A 9.879311 > 
AB 2.519828 \ 


sin. 90° 


10.000000 


AB 

AC, 506.91 


2.519828 


12.399139 


2.704928 



CB, 383.92 2.584239 

Ex. 6. Given AG, 45, and the angle C, 37° 22', to find 
the other parts. 



64 KEY TO [£96 

The angle A must be 52° 38'. 

For GB. For AB. 

As sin. 90° 10.000000 sin. 90° 10.000000 

Sin. G, 37° 22' 9.783127 sin. A 9.900240 

AG 1.653212 AG 1.653212 



AB, 27.311 1.436339 GB, 35.764 1.553452 

Ex. 7. Given AG, 4264.3, and the angle A, 56° 29' 13", to 
find the other parts. 

The angle G must be 33° 30' 47". 

For GB. For AB. 

As sin 90 D 10.000000 sin. 90° 10.000000 

Sin. A, 56° 29' 13" 9.921041 sin. G 9.742038 

AG 3.629848 AG 3.629848 



GB, 3555.4 3.550889 AB, 2354.4 3.371886 

Ex. 8. Given AB, 42.2, and the angle A, 31° 12' 49", to 
find the other parts. 

The angle G must be 58° 47' 11". 

For BG. For AG. 

As sin. G, 58° 47' 11" 9.932088 sin. 9.932088 

Sin. A, 31° 12' 49" 9.714522 sin. 90° 10.000000 

AB 1.625312 AB 1.625312 



11.339834 AG, 49.34 1.693223 



BG, 25.57 1.407746 



Ex. 9. Given AB, 8372.1, and BG, 694.73, to find the 
other parts. 
(By Prop. III. Plane Trig., 2d part, we have,) 
8372.1 : 694.73 : : B : tan. A. 



£90] PLANE TRIGONOMETRY. 65 

(694.73)i2 log. 12.841816 



Whence, tan. A 



8372.1 log. 3.922835 



tan. A = tan. 4° 44' 37"= 8.918981 
Therefore the angle (7=85° 15' 23" 

-For the hypotenuse we have, 

As sin. A : 694.73 : : B : AG 
As before. log. (694.73)i2= 12.841816 

Sin. A 8.917489 

AC, 8400.9 3.924327 

Ex. 10, Given AB, 63.4, AC, 85.72, to rind the other 
parts. 

As AG : sin. 90° : : AB : sin. C 





Log. (ABM) 


11.802089 




Log. AC, 85.72 
Sin. (7=47° 41' 56 !f 


1.933082 




9.869007 


Again, 


As B 


10.000000 




Is to AG 


1.933082 




Sin. A— cos. G 
Log. BG= log. 57.69 


9.828032 




1.761114 


Ex. 11. 


Given AG, 7269, and AB, 3162, to find th 


parts. 






As in Example 10, Log. {AB.B) 


13.499962 




Log. AG, 7269 


3.861475 




Sin. G, 25° 47' 7" 


9.638487 




A 64° 12' 53" 




As 


sin. C, 25° 47' 7" 


9.638487 


Is to 


^£, 3162 


3.499972 


So is 


sin. ^4= cos. G 


9.954450 




Log. £<7=log. 6545 
5 


13.454422 




3.815935 



66 KEY TO [£94>, §91 

Ex. 12. Given AG, 4824, and BG, 2412, to find the other 
parts. 

In this example the hypotenuse is double of the side 
BG, therefore the angle opposite BO or A, must be 30°, and 
the angle G, 60°. 

If we call BG, 1, AG, 2, and AB, x, then a 2 + 1=4, and 
x= VI. Whence, ^5=2412^3=4178, nearly. 

Ex. 13. In this example the hypotenuse of the right an- 
gled triangle is 94,770,000 miles, and the most acute angle 
only 16' 6", the double of the side opposite this small angle 
is required. 

The difficulty here is to obtain the sine of 16' 6" to a 
sufficient degree of accuracy from the common table. This 
is explained on page 288, Text-book. Or we may operate as 
follows, 

As Sin. 90° 10.000000 

Is to 94,770,000 log. 7.976671 

So is Sin. 16' 6"= (16.1) sin. 1' 6.463726, sin. 1' 

Log. 16.1 1.206826 

443850 5.647223 

2 



Sun's diameter 887700 

Ex. 14. Solution, As 94,779,000 7.976671 

R 10.000000 

Los. 3956 3.597256 



Sine of solar parallax= 5.620585=8.61". Ans. 

Ex. 15. The distance from the earth to the moon, at 
mean distance in miles, is (60.3) (3960). 



PLANE TRIGONOMETRY. 



67 



Ml] 

Statement, As sin. 90° : sin. 15' 32" : : (60.3)(3960) : M, 

Observe that 15' 32" = 932", sin. 932"=932 times sin. 1" 

Solution. 
Log. 60.3 1.780317 

Log. 3960 3.597695 

Sin. 1" (see page 288, Geom.) 4.685575 
Log. 932 2.969416 

Sum, less 10, B, 1078.9 3.033003 

2 

Diameter of moon, 2157.8 Ans. 



OBLIQUE ANGLED TRIGONOMETRY. 



(practical problems.) 



(Page 295.) 



Note. — One triangle will serve for the solution of nearly all the following 
examples in this section. But the learner should draw his figure as near as he 
can according to the data given in each example. He will thus be less likely 
to make mistakes than he otherwise would. 

Let ABC be any oblique angled tri- 
angle. 

Ex. 1. Given AB, 697, the angle 
A, 81° 30' 10", and the angle B, 40° 
30' 44", to find the other parts. 

Because the sum of the three angles make two right an- 
gles, or 180°. Therefore, 

<7=180°-(81° 30' 10" + 40° 30' 44")==57° 59' 6". 

As Sin. C : AB : : sin. A : BC. 




68 



KEY TO 



[295 



Whence, 


To sine A, 81° 30' 11" 


9.995207 




Add, 697 


2.843233 




12.838440 




Sub. sin. G, 57° 59' 6" 


9.928350 




BC, 813 log. 


2.910090 


Again, 


Sin. : AB : : sin. 5 


: AG 




Log. ^5 


2.843233 




Sin. B 


9.812652 
12.655885 




Sin. C, as before 
AG, 534 


9.928350 




2.727535 



Ex. 2. Given AC, 720.8, the angle A, 70° 5' 22", and 
B, 59° 35' 36", to find the other parts. 
The angle G must be 50° 19' 2". 

AG. sin. A 



As 
Also, 



Sin. B 
Sin. B 



AG 



sin. A : BG= 



Sin. ^ 
u4(7, 720.8 

Sin. B 
BC, 785.8 



AG : : sin. (7 : AB= 

9.973232 sin. G 

2.857815 



12.831047 
9.935737 



sin. B 



Sin. B 
A C. si n. (7 
Sin. 5 ' 
9.886260 
2.857815 
12.744075 
9.935737 



2.895310 AB, 643.2 2.808338 



Ex. 3. Given 5(7, 980.1, the angle A, 7° 6' 26", and the 
angle B, 106° 2' 23", to find the other parts. 
The angle C must be 66° 51' 11". 

BC. sin. 5 



Sin. A : BC : : sin. 5 : AG-- 
Sin. A : BC : : sin. (7 : ^5= 



Sin. ^ 
£<7. sin. C 



Bin. A 



£95] PLANE TRIGONOMETRY. 69 



Sin. B 


9.982755 


sin. G 


9.963552 


BG 


2.991270 


BG 


2.991270 




12.974025 


12.954822 


Sin. A 


9.092463 


sin. A 
AB, 7283.8 


9.092463 


AC, 7613.1 


3.881562 


3.862359 



Ex. 4. Given AB, 896.2, BG, 328.4, and the angle 
O, 113° 45' 20'' to find the other parts. 

As AB : sin. G : : BG : sin. A= B0 ' si g-?. 

AB 

Sin. (7= cos. 23° 45' 20" 9.761550 

BG, 328.4 2.516403 

12.477953 

AB, 896.2 2.952405 

Sin. .4= sin. 19° 35' 46" 9.525548 

Now having the angles A and G, we subtract their sum 

from 180°, and thus find the remainder B, 46° 38' 54". Ans. 

With this angle we determine A G 



Ex. 5. Given AG=4627, BG, 5169, and the angle A, 
70° 25' 12", to find the other parts. 

As BG : sin. A : : AG : sin. ff=^, sin - ^ 

Sin. A, 70° 25' 12" 9.974132 

AG, 4627 3.665299 

13.639431 
BG, 5169 3.713407 

Sin. £ ; 57° 29' 56" "^926024 

We now have the angles A and B, subtracting their sum 
from 180°, gives fc52° 4' 52". Ans. 



70 



\ 




KET2 


■ TO 




[296 


For AB, we have, 










Sin. 


A : BC : : 


sin. 


c : .lb-* : sin : 

sin. A 


a 




BC 






3.713407, 






Sin. 


<7, 52° 4' 52' 


/ 


9.897011 








13.610418 




Sin. 


A 
4328 




9.974132 






AB, 


3.636286 





Ex. 6. Given -45, 793.8, BC, 481.6, ^4(7, 500, to find 
the angles. 

Note. — Here all three sides are given, and the solution is 
by formulas found on 'page 259, Geometry. 

The angles of a triangle are denoted by A, B, C, and the 
sides opposite by a, b, c. A opposite a, etc. 

The formulas are ±(a + b-\-c)=S. 

be ac • ab 

the radius being unity. 

By logarithms the formulas become 

Log. (cos. 9 ^)=log. S(S— a)— log. be. &c. 
a= 481.6 Log. S 2.948266 

c b= 500 Log. (S-a) 2.608633 * 

I c= 793.8 Log. S(S-a) '5.556899 



> sub. 



2)1775.4 log. b, 2.698970 

887.7 =S log. c, 2.899711 

a, 481.6 5.598681 5.598681 J 

406.1 = (S—a) log. (cos. 2 i^) -1.958218(2, div. 

log. (cos. \A) -1.979109 
To correspond to table II., add 10. 

Cos. ^=cos. 17° 37' 46" 9.979109 

2 



A= 35° 15' 32". 



296] PLANE TRIGONOMETRY. ?J 

Having thus found one of the angles, the other may be 
determined by the direct proportion between the sides and 
the sines of the opposite angles. 



Ex. 7. Given AB, 100.3, AG, 100.3, and BO, 100.3, to 
find the angles. 

Here the sides are equal to each other, the triangle is there- 
fore equiangular as well as equilateral. Consequently each 
angle is one third of 180°, or 60°. 

Ex. 8. Given AB, 92.6, BG, 46.3, and AG, 71.2, to find 
the angles. 

Note. — In example 6 we obtained the angle A, for radius 
unity. Here we will find the angle B, for radius B, whose 
log. is 10. The formula must be 



co S . ,B=m^m. 

\ ac J 



By logarithms this becomes, 

Log. (cos. i£) = i(log. R 2 S(S-b) -log. ac). 



b= 71.2 

a= 46.3 

c= 92.6 

2)210.1 

105.05=/? 


-b. 

Jos. 


log. a, 
log. c, 

±B=cos, 
B= 


Log. JB a : 
log./? 
log. (S- 

1.665581 

1.966611 


= 20 

2.021396 
-b) 1.529559 

23.550955 


71.2 


3.632192 

. 24° 23' 45" 

2 


3.632192 


33.85=/?- 
C 


2)19.918763 
9.959381 




48° 47' 30". 





72 



KEY TO 



[296 



Now, 



sin. B 



sin. A = 



a sin. B 





Log 


. a— 




1.665581 




Sin. 


B= 




9.876402 
11.541983 




Log. b = 

Sin. A = sin. 29° 


1.852480 




17' 22" 9.689503 


Hence, 




C= 


101° 


55' 8" 


Ex. 9. 


Given 


AB, 


4963, 


BC, 5124, AC, 5621, to find 


le angles. 








a- 


= 5124 




• J 


'or the angle C 


b~- 


=5621 






log. JB 9 20. 


c- 


=4963 




log. 


log. S 3.895091 




15708 


log. (8-c) 3.461048 


S= 


7854 


a, 3.709609 27.356139 


S-c= 


2891 




log. 


b, 3.749814 



7.459423 



cos. i(7=cos. 27 D 23' 27.5" 



7.459423 

2)19.896716 

"9.948358 



C-- 

c : sin. C 
Sin. C 
Log. a 



i xw 



54° 46' 55 

For the angle A 
a : sin. A. 



Log. c 

Sin. A= sin. 57° 30' 28" 



9.912202 
3.709609 
13.621811 
3.695744 
9.926067 



Ex. 10. Given AB, 728.1, BC, 614.7, AC, 583.8 to find 



the angles. 



296] PLANE TRIGONOMETRY. 73 

a=614.7 log. B 2 20.000000 

5 = 583.8 j log. 2.766264 log. 8 2.983762 

c=728.1 ) log. 2.862191 log. (8— a) 2.542327 

192&6 5.628455 25.526089 

S 963.3 5.628455 



614.7 2)19.897634 

(tf-a) 34a6 cos. J.4=cos. 27° 16' 26" 9.948817 

2 



^ 54° 32' 52" 

■kt a x • ■ r> ^ sin. ^4 

JNowas a : sm. ^4 : : 6 : sin. i?= , 

a 

log. 5 2.766264 

sin. A 9.910944 



12.677208 

log. a, 614.7 2.788663 

sin. £=sin. 50° 40' 58 J < 9^888545 



Ex. 11. Given AB, 96.74, BG, 83.29, and AC, 111.42, 
to find the angles. 

a= 83.29 log. B 2 20.000000 

5=111.42 ) log. 8 2.163534 

c= 96.74 \ log. (S-a) 1.795428 

2)291.45 log. b, 2.046963 2&958962 

8 145.725 log. c, 1.985606 

a 83.29 4032569 4.032569 

{S-a) 62.435 2)19.926393 

Cos. J^ = cos. 23° 15' 22.5" = 9.963196 



A = 46° 30' 45" 



74 


KEY TO 


As 


a : sin. A : : b : sin. i? 




Sin. A = 9.860652 




I 2.046963 




11.907615 




a 1.920593 




Sin. B, sin. 76° 3' 46" 9.987022 



[£96 



A and B being as above, C must be 57° 25' 29". 

Ex. 12. Given AB, 363.4, BO, USA, and the angle 
B, 102° 18' 27", to find the other parts. 

Here we have two sides and their included angle given, and 
we apply (Prop. VII., Plane Trig., page 257), which is 
363.4) 363.4 } (C+A\ , (C~A\ n . 

USA ! SUm ' : 148.4 ( dlff ' : : ^ (-2-) : tan ("2-J (1) 

We observe that O must be greater than the angle A, be- 
cause O is opposite the greater side. 



As 



511.8 


215 : : 


tan. 38° 50' 47" : tan. — -*. 

A 




Log. 


215 


2.332438 




Tan. 


G+A , 38° 50' 47" 


9.905986 




12.238424 




Log. 


511.8 


2.709100 




Tan. 


\(C-A) 18° 41' 29" 

i(<7+^)38°50'47" 


= 9.529324 




Sum 


O 57° 32' 16" 


i Ans. 




Diff. 


A 20° 9' 18" 


3 


Sin. 


C : c(363.4) : : sin 


,B:b. 


Sin. B, 102° 18' 27"=cos. 12° 18' 27" 


9.989903 






log. 363.4 


2.560385 




12.550288 






sin. C, 57° 32' 16" 


9.926211 



loa;. AC=lo%. 420.8 2.624077 




£96] PLANE TRIGONOMETRY. 75 

Ex. 13. Given AB, 632, BG, 494, and the angle A, 
20° 16', to find the other parts, the angle C being acute. 

,C If no mention were made requiring 
the angle G to be acute, the data 
would give the triangle ABC, for 
the solution as well as the triangle 
ABG. 

In such cases it is customary to solve both triangles, and 
call the solution ambiguous. 
As, BG : AB : : sin. A : sin. G 

Or, 494 : 632 : : sin. 20° 16' : sin. G 

Log. sin. 20° 16' 9.539565 

log. 632 2.800717 

12.340282 
log. 494 2.693727 

G, 26° 18' 19" 9.646555 

A and G taken from 180°, gives 133° 25' 41", for B. 
Lastly for AG, we have, 

Sin. A : BG : : sin. B : AG (b). 
Sin. B, 133° 25' 41"=cos. 43° 25' 41" 9.861079 

log. BG 2.693727 

12.554806 
sin. A 9.539565 



AC, 1035.7 3.015241 

Ex. 14. Given AB, 53.9, AG, 46.21, and the angle B, 
58° 16' to find the other parts. 
As, 46.21 : sin. B, 58° 16' : : 53.9 : sin. G 

Sin. B, 58° 16' 9.929677 

AB, 53.9 1.731589 

11.661266 
AC 46.21 1.664736 



Sin. G, 82° 46' 9.996530 



76 ' KEY TO [2®7 

B and G being known, their sum, 141° 2' taken from 180°, 
gives 38° 58', for the angle A. Ans. 

As, sin. G : AB, 53.9 : sin. A, 38° 58' : BG 

Sin. 38° 58' 9.798560 

AB, 53.9 1.731589 

11.530149 
Sin. (7, 82° 46' 9.996530 



BG, 34.16 1.533619 

Ex. 15. Given AB, 2163, BG, 1672, and the angle G, 

112° 18' 22", to find the other parts. 

As, sin. G : AB, 2163 : : sin. A : BG, 1672. 

ttti cr ^ 1672 sin. (7 
Whence, Sin. .4 = 2163""' 

Sin. (7= cos. 22° 18' 22" 9.966221 

log. 1672 3.223236 



13.189457 



log. 2163 3.335057 

Sin. A, 45° 39' 22" 9.854400 

Now, A and (7 being known, .£ must be 22° 2' 16". 
Sin. A : 1672 : : sin. B, 22° 2' 16" .: AG. 
Sin. 22° 2' 16" 9.574283 

BG, 1672 3.223236 

12.797519 
Sin. A, 45° 39' 22" 9.854400 

AG, 877.2 2.943119 

Ex. 16. Given AB, 496, £(7, 496, and the angle B, 
38° 16' to find the other parts. 
In this example we observe that AB—BG, therefore, the 



£97] PLANE TRIGONOMETRY. 77 

angles A and C, must be equal to each other, and the value 
of each must be 

i(180°-38°16')=70°52'. 
Now, Sin. 70° 52' : 496 : : sin. 38° 16' : AC. 

4r _ 496.sin. 38° 1 6' 
sin. 70° 52' ' 

Sin. 38° 16' 9.791917 

Log. 496 2.695482 

12.487399 
Sin. 70° 52' 9.975321 



AC=325.1 2.512078 

Ex. 17. Given AB, 428, the angle C, 49° 16', and (AC+ 
CB) = 918, to find the other parts. 

Note. — This problem is the same as 30, in Book VIII., 
Geometry, page 242, and its general solution is given in this 
Key, on page 59. 

D 

Either triangle, ABC or AB' O will corres- -^ 

pond with the data. We will take the trian- CX 

gle ABC, and commence by solving the ®/ 

triangle ABB. ^^\ 

Because CB=CB, the angle ^4 (7i? is double A ^^^^ 
of the angle D. 
Therefore, D=24° 38', AD=918, AB=428. 
As 428 : sin. D : : 918 : sin. ABB. 

Log. 918 2.962843 

Sin. 24° 38' 9.619938 

12.582781 
Log. 428 2.631444 



Sin. ABB\ 63° 22' 48" 9.951337 



78 





KEY TO 


[297 


Or, 


ABD= 


116° 37' 12" 


Sub. 


CBD= 
ABC= 


24° 38' 00 


DifT.= 


91° 59' 12" * 



To ABC, add ACB, 49° 16', and their sum, 141° 15' 12", 
taken from 180° gives .4=38° 44' 48". 

Lastly, Sin. C : AB : : sin. A : BC. 
Which is, Sin. 49° 16' : 428 :: sin. 38° 44' 48" : BC. 
Sin. 38° 44' 48" 9.796490 

Log. 428 2.631444 

12.427934 
Sin. 49° 16' 9.879529 



BC, 353.5 2.548405 

Now, as (AC+BC)=918, we have ^(7=918-353.5= 
564.5. In like manner, solving the triangle ABO, we shall 
have AC>=353.5, and j5'G"=564.5. 

Ex. 18. Given a side and its opposite angle, and the dif- 
ference of the other two sides, to construct the triangle and 
find the other parts. 

Let ABC be the triangle. AC=126, 
B=29° 46', and AM, the difference be- 
tween AB and BC, =43. 
>B From 180° take 29° 46' and divide the 
remainder by 2. This gives the angle BMC or BCM. BMC 
taken from 180°, gives AMC. 

Now in the triangle AMC, we have the two sides AC, 
126, AM, 43, and the angle AMC, to find the angle A. 
The computation is as follows: 180° -29° 46' = 150° 14' ; 
half, =75°7'=BMC 180° -75° 7' =104° 53'= AMC. Now 
in the a AMC, we have 

AC : AM : : sin. 104° 53' : sin. ACM 




<J97] PLANE TRIGONOMETRY. 79 

126 : 43 : : cos. 14° 53' : sin. ACM 
Cos. 14° 53' 9.985180 

Log. 43 1.633468 

11.618648 
Log. 126 2.100371 

Sin. AOM=sin. 19° 15' 28" 9.518277 
Now to BCM, 75° 7' add ACM, 19° 15' 28", and we have 
ACB, 94° 22' 28". Subtracting 19° 15' 28" from 75° 7', that 
is, ACM from BMC, and the difference must be 
^=55° 51' 32". 
Lastly, As sin. B : AC, 126 : : sin. A, 55° 51' 32" : CB. 



Sin. A 


9.917851 


126 


2.100371 




12.018222 


Sin. B, 29° 46' 


9.695892 


BC, 210.05 


2.322330 



Ex. 19. Given AB, 1269, AC, 1837, and the including 
angle A, 53° 16' 20", to find the other parts. 
Solution by the same formulas as in 12. 

I: S i sum •• 2S i diff - : : tan - ^ +c > ■■ tan - ^ B -v- 

3106 : 568 : : tan. 63° 21' 50" : tan i(B-C). 

Tan. 63° 21' 50" 10.299685 

Log. 568 2.754348 

13.054033 
Log. 3106 3.492201 

±(B-C), 20° 1'57" 9.561832 

But K^+O)=63°21'50» 

Sum= 5=83° 23' 47". Diff. C=43° 19' 53". 



80 - KEY TO [305 

Lastly, Sin. B : b : : sin. A : a(BC) 

Sin. A, 53° 16' 20" 9.903896 

Log. b, 1837 3.264109 

13.168005 

Sin. B, 83° 23' 47" 9.997109 

AC, 1482.16 3.170896 



SECTION III. 

APPLICATION OF PLANE TRIGONOMETRY. 
(Page 305.) 

(1.) Kequired the height of a wall whose angle of eleva- 
tion, at the distance of 463 feet is observed to be 16° 21'. 

This example presents a right angled triangle, the base 
of which is 463, and the acute angle at the extremity of the 
base 16° 21'. 

Let x= the required height ; then 

7? * irooii acq 463 tan. 16° 21' 
B : tan. 16° 21' : : 463 : x— ^ . 

Log. 463 2.66b581 

Tan. 16° 21' 9.467413 



Ans. 135.8 2.132994 



(2.) We solve this example by the adjoining geometrical 



figure. 




PLANE TRIGONOMETRY. gl 

ABC is a triangle ; CBD its exte- 
rior angle. 

Whence, CBD=ACB+A. 
-A That is, 31°18'=^<7£ + 26°18', 
or, ACB=5°, AB=2U. 

Sin. 5° : 214 : : sin. 26° 18' : CB. 

Whence, BO=^^^^L 

sm. 5 

Again, in the triangle CBD we have 

Sin. 90° : CB : : sin. 31° 18' : CD. 

That is, Sin. 90° : 2 l- sin> 26 ° — : : sin. 31° 18' : CD. 

sin. 5° 

™ ^ n 214.sin. 26° lS'.sin. 31° 18' 

Whence, CD = : — — _, . — — . 

sm. 90 . sin. 5 

214 2.330414 

Sin. 26° 18' 9.646474 

Sin. 31° 18' 9.715602 

Log. Num. 21.692490 
Sin. 90° 10.000000 

Sin. 5° 8.940296 18.940296 



CD=565.2 2.752194 

(3.) Here the perpendicular of a right angled triangle is 
given 149.5 feet, and the vertical angle 57° 21'. The oppo- 
site angle is therefore 32° 39'. 

As, Sin. 32° 39' : 149.5 : : sin. 57° 21' : Ans. 

Or, As Cos. 57° 21' : 149.5 : : sin. 57° 21' : Ans. 

Ans. = U9.5. s iHl^_^ = 149.5. tan. 57° 21'. 
cos. 57° 21' 

149.5 2.174641 

tan. 57° 21' 10.193307 



Ans. =233.3 2.367948 

6 



82 



KEY TO 



[305 



(4.) Here are two right angled triangles to be solved. 
The angle of depression, is equal to the angle opposite to 
the perpendicular. 

Hence, for the distance of the mean object, we have the 
following proportion. 

Sin. 48° 10' : 138 : : cos. 48° 10' : Dis. 

Dismiss. 008 ' 48010 ' — ■ — 



"" ~~'sin. 48° 10' 






* 


138 






2.139879 


cot. 48° 10' 






9.951896 


Least dis. 123.52 






2.091775 


The greatest dis. = 138.cot. 


18° 


52' 





(5.) Here is but one acute angled plane triangle, and the 
angle opposite to the base is found thus. 

31° 15' 

86° 27' 



117 42 
180 
62° 18' 



Let x represent the distance from one extremity of the 
base to the house, and y represent the distance of the other 
extremity. Then 



a=312 

312 
Sin. 31° 15' 

Sin. 62° 18' 
x= 182.8 



Sin. 62° 18' : 312 
Sin. 62° 18' : 312 
sin. 31° 15' 



: sin. 31° 15' : x 
: sin. 86° 27' : y. 
„,„ sin. 86° 27' 



sin. 62° W 
2.494155 
9.714978 
12.209133 
9.947136 
2.261997 y=351.7 



u ~ "; sin. 


62° 18'' 


312 


2.494155 


sin. 86° 27' 


9.999166 




12.493321 


sin. 62° 18' 


9.947136 



2.546185 



3©6] PLANE 


TRIGONOMETRY. 83 


(6.) This is in many 

nnm nova 


respects 


the 


same as (5), except in 


ilUIii Ucl b. 

Sin. 60° : 


260 :: 


sin. 


40° 


: x. 


Log. 260 








2.414973 


Sin. 40° 








9.808067 
12.223040 


Sin. 60° 








9.937531 


192.8 


2.285509 



192.8 is the distance from one extremity of the base to the 
tree, but this line (192.8) makes an angle with the base of 
80°. Let 192.8 be a hypotenuse of a right angled triangle, 
and the angles are 80° and 10°. 

The side opposite 80° is the line or distance required. 
Sin. 90° : 192.8 : : sin. 80° : perpendicular. 
192.8 2.285509 

Sin. 80° 9.993351 

Ans. 190.1 2.278860 



(7.) Let BC be the eminence, 268 feet, 
and AD the steeple. Draw CE parallel 
to the horizontal AB. Then ECD=40° 3', 
ECA = CAB =56° IS 1 . 

DC A =56° 18' -40° 3' = 16° 15', DAC= 
90°-56° 18' = 33° 42'. 
In the A ABO, we have 

Sin. 56° 18' : 268 : : sin. 90° 
268 xB 
sin. 56° 18'' 

In the a ADC, we have the supplement to the angle 
ADC, equal to 16° 15' added to 33° 42', or 49° 57 
fore, 




' ; there- 



As 



Sin. ADC : AC : : sin. DC A : AD. 



84 



That is, Sin. 49° 57 



KEY TO 

268 xB 



[306 

sin. 16° 15' : AD. 



' smT5"6°l8' 
Log. AD = 

log. [268.72.sin. 16° 15'] -log. [sin. 49° 57'.sin. 56°18'] = 
[2.428135 + 10 + 9.446893] - [9.883936 + 9.920099] = 
21.875028-19.804035=2.070993 • 
Whence, AD=I17.76 feet. 




(8.) Let C and D be the two objects, 
and A the point at which both can be 
seen. 

AD =1428, A (7=1840, and the an- 
gle at A 36° 18' 24" 

From 180° 



Angles C and X>=143° 41' 36" 
i sum 71° 50' 48" 

Here we will apply the following theorem in trigonometry. 
As the sum of two sides is to their difference, so is the 
tangent of half the sum of the angles at the base, to the 
tangent of half their difference. 

Let x— the half difference between D and O. 



Then, 




3268 : 412 








Or, 




817 : 103 : 


tan. 


71 G 


50' 48" : tamo;. 




n 


Log. 103 

Tan. 71° 50' 48" 

817 
tan. 21° 1'55" 
71° 50' 48" 






2.012837 
10.484284 




12.497121 
2.912222 


Tan. a 


9.584899 


Angle 
Angle 


D 

C 


= 92° 52' 43" 
50° 48' 53" 









306] 



PLANE TRIGONOMETRY. 



85 



Sin. 50° 48' 53" : 1428 : : sin. 36° 18' 24" : CD. 
1428 3.154728 

Sin. 36° 18' 24" 9.772400 



Sin. 50° 48' 53" 
Ans. 1090.85 

(9.) Let AB represent the mountain, 
and AD the visible distance. AB pro- 
duced will pass through the center of the 
earth at C. From D draw CD perpen- 
dicular to AD. Join BD. ADC is a 
right angled triangle. 

CAD=90°-2° 13' 27" = 87° 46' 33". 
ACD=2° 13' 27". ADB=z\ACD = l° 6' 
43.5". XSZ>=91°6'43.5". 
Now in the A ABD, we have 

Sin. 1° 6' 43.5" : 3 : : sin. 91° 6' 43. 
Sin. 91° 6' 43.5" = cos. 1° 6' 43.5" 
Log. 3 



12.927128 

9.889362 
2.037766 




5" : AD. 
9.999919 
0.477121 





10.477040 


Sin. 1° 6' 43.5" 


8.287976 


Log. AD, 154.54 


2.189064 


In the triangle ADC, we have 




sin. ACD : AD : : cos. 


ACD : CD. 


Cos. A CD= cos. 2° 13' 27" 


9.999674 


AD 


2.189064 




12.188738 


£in. ACD=sin. 2° 13' 27" 


8.588932 


CD 


3.599806 


Log. 2 


0.301030 



Diameter, log. 7958 miles, 



3.900836 



86 



KEY TO 



[306 



(10.) Let H be the location of the headland, A the posi- 
sition of the ship when the first observation was taken, and 
B its position when the last observation was taken. 

AB =20 miles, and on A, mB, meridians passing through 
the eye of the observer at the two stations. 

The data gives us the ' angle mAB 
=47° 49', mBH=87° 11', the sum of 
these two angles taken from 180°, 
gives the angle ABH=45°. 

The angle BAH, is obviously equal 
to the sum of 47° 49', and 39° 23', 
which is 87° 12'. The sum of 87° 12' and 45° taken from 
180° gives AHB=4J° 48'. 

We now have AB=20 miles one side, and each of the 
angles, A, B, and H, to find AH, and BH. 




For AH 


Sin. 47° 


48' : 


20 : : sin. 45° 


: AH 


For BH, 


sin. 47° 48' 


: 20 


: : sin. 87° 12' : 


BH 


Sin. 45° - 


9.849485 




sin. 87° 12' 


9.999481 


Log. 20 


1.301030 






1.301030 




11.150515 


11.300511 


Sin. 47° 48' 


9.869704 
1.280811 




BH, 26.96 


9.869704 


AH, 19.09 


1.430807 



(11.) By (Th. XVIII., Book III.), the length of a line 
drawn from the top of the tower to touch the surface of the 
sea, must be, V 79 60. 5280. 100, and from the same point on 
the sea, the line extended to the mast head of the ship must 
be 4/7960.528090. The problem requires the sum of these 
two lines. 

The computation by logarithms is as follows, 



S06] 

Log. 7960 
Log. 5280 
Log. 100 



PLANE TRIGONOMETRY. 

3.900913 
3.722634 
2.000000 



log. 90 



87 

3.900913 
3.722634 
1.954243 



2)9.623547 
64829.6 4.811773 

Sum in feet, 126332.4 

Log. 5280 

Sum in miles, 23.92 

Add T \ for refraction 1.84 



2)9.577790 
61502.8 4.788895 
log. 5.101514 
3.722634 



1,378880 



25.76 Ans. 



(12.) As 


Sin. 35° : 143 feet : 


: cos 


35° : Dis 


Whence, 


Dist. =143. C ?^-?^ 
sm. 35° 


= 143.cot. 35°. 




Log. 143 




2.155336 




Cot. 35° 

Ans. 204.22 




10.154773 




2.310109 



(13.) Let CD be the breadth of the 
river, or the distance sought, ^J5=500 
yards, the measured base. 

The angle ABC=53°, and BAG, 79° 
12'. Whence the angle A OB = 47° 48'. 

Now, 
Sin. 47° 48' : 500 : : sin. 79° 12' : BC. 

™ r> n 500.sin. 79° 12' . 

Whence, BC— —. — Ano AOI — 

' sm. 47° 48' 

Again in the right angled triangle BBC, we have 
Sin. D : BC : : sin. B : DC. 




88 KEY TO [307 

That is, Sin. 90 : BC : : sin. 53° : DC 

wu -nn 500.sin. 79° 12'.sio. 53° 

Whence. X>C/=- : — ._. .„, . — 7 r^. 

; sm. 47 J 48'.sin. 90° 

Log. 500 2.698970 

Sin. 79° 12' 9.992239 

Sin. 53° 9.902349 



22.593558- numerator, 
Log. B, sin. 47° 48' 19.869704 denominator. 
Log. OZ>=log. 529.48 2.723854 



(14. ) Let L represent the length 
of the inclined plane, and P its 
perpendicular height. 

The angle <7J5D=46°, and CAB 
B * =31°. Whence, ACB=15°, and 

^4P=200. The triangle ABC gives the proportion 

Sin. 15° : 200 : : sin. 31° : x= 200.sin. 31° 

sm. 15 u 

From the triangle CBD, we obtain, 

Sin. 90° : L : : sin. 46 3 : P= L ' silL f° 

sm. 90° 

mux- t> Avn /200.sin. 31°\sin. 46° 

That is, P or CZ>=( , — ^ -. — ^o- 

V sm. 15°/sm. 90° 




Log. 200 2.301030 

Log. sin. 31° 9.711839 

Los. sin. 46° 9.856934 



Log. of numerator 21.869803 

L02. of denominator 19.412996=P.sin. 15° 



Log. 286.28 2.456807 

(15.) Take the same triangle as in the preceding example. 



307] 



PLANE TRIGONOMETRY 



89 



UW J ■*■ 


i^.sin 


.26° 


jB.sin. 26° 


Log. 300 




2.477121 


2.477121 


Sin. 32° 




9.724210 


9.724210 


Sin. 58° 




9.928420 


cos. 58° 9.724210 


Log. num. 


22.129751 


21.925541 


Log. denom. 


54 


19.641842 


19.641842 


P=307. 


2.487909 


dis. 192.18 2.283699 



(16.) Here we have a triangle, one side of which is 440 
yards, and the adjacent angles, 83° 45', and 85° 15', there- 
fore the angle opposite must be 11°. 

Now, for the side of the triangle which is opposite the 
angle 85° 15', we have the following proportion. 
Sin. 11° : 440 : : sin. 85° 15' : x. 
For the other side, we have 

Sin. 11° : 440 : : sin. 83° 45' 
For x. 
2.643453 
9.998506 



y- 



Log. 440 
Sin. 85° 15" 



sin. 83° 45' 



For y. 
2.643453 
9.997411 



Sin. 11° 
2298.05 



12.641959 
9.280599 
3.361360 



12.640864 
9.280599 



2292.26 3.360265 



(17.) Let A and B be the 
positions of the ship, when the 
observations were taken. 

Then ^4.5=12 miles in the 
direction north east. 

Then, also AL, is the direc- 
tion to the land, and BL, is 
another direction to the same point. 




90 



KEY TO 



[308 

The angle LAB is 5 points of the compass, or 56° 15', 
and the angle ABL, is 9 points, or 101° 15'. 

Hence, the angle at L must be 22° 30'. 
Now, Sin. 22° 30' : 12 : : sin. 56° 15' : BL. 

12sin. 56° 15' 



BL = 

Log. 12 
Sin. 56° 15' 

Sin. 22° 30' 
Ans. 26.072 



sin. 22° 30' 



1.079181 
9.919846 
10.999027 
9.582840 
1.416187 



(18.) This problem requires the 
adjoining figure. 

We must compute the angle CAB, 
to find BAO. 

And from the triangle ABD, we 
must compute the angle ABD, to 
find ABO. 
We compute the angle CAB, by the following formula 
(Prop. 8, Plane Trigonometry.) 




Cos . u=v2£=&. 

DC 

For the triangle ABC, 

a=560, 5 = 100, c = 500. 
Whence, s=580, and s— a =20. 



Cos ■j-,/^580x-20 /WUB 



307] PLANE TRIGONOMETRY. 91 

Log. E 2 116 22.064458 

Log. 500 2.698970 

2)19.365488 
Cos. \A= 61° 12' 21"= 9.682744 
2 



A 122° 24' 42" 
180° 



BAO= 57° 35' 18" 

In the triangle ABJD, changing D to O, then to find the 
angle B, we have 



Cos. ±B=\/ R2s ( s *)• a formula in which 
r ac 

a=100, 5=550, c=500. 

Whence, s=575, and s—b=25. 

n tT> A /B 2 575x25 ./^Tl^ 

cos. i5= V - 1M -^ m =V- m - 

B*115 22.060698 

400 2.602060 



2)19.458638 



Cos. |J5=cos. 57° 34' 31" 9.729319 
2 



115° 9' 2" 
180 



ABO 64° 50' 58" 

BAO 57° 35' 18" 



122° 26' 16" 
180° 



Angle 0= 57° 33' 44" 
Sin. 57° 33' 44" : 500 : : sin. 57° 35' 18" : 50=500.14 



92 



KEY TO 



[308 




(19.) Let AQB be 45 minutes of 
a degree, although in the figure be- 
fore us, it is many degrees. 

Let A be the position of the eye 
of the observer, and conceive the 
line A By to touch the earth at t. 
Then the angle ACt=4! 15". 
Hence, tCH=4W 45". 

Join tH. 

Now the angle made between the chord tH and the tan- 
gent tB, is measured by half the arc, therefore, the angle 
BtH=20 1 22i". 

The angle DAB is 31' 20", and BD is the visible part of 
the mountain, and BH is the invisible part. We must first 
compute the tangents At, and tB. 

R : tan. 4' 15" : : 3956 : tA 
tan. 40' 45" : : 3956 : tB 
3.597256 



B : 



Log. 3956 
Tan. 4' 15" 

At 4.8907 



7.092115 
0.689371 



-tan. 40' 45" 
tB 46.8954 
At 4,8907 



3.597256 
8.073874 

1.671130 



AB 51.7861 



* The sine or tangent of an arc exceeding 20' and less than 2° can not 
be computed accurately by the common method nor by the method explained on 
page 288, (Plane Trigonometry.) 

In the present case we may employ the familiar formula. 
Sin. 2a=2 sin. a.cos. a. 
a=(40'45")-i-8=5' 5.625". 



Put 

Then, we have 



Log. 2 .301030 

Sin. 5' 5.625// T.l'70'764 (See page 288, Trig.) 

Cos. 5' 5.625" 10.000000 

7.47l794=sm. 10' 11.25/'. 



308] 



PLANE TRIGONOMETRY 



93 



In the triangle tBH, we have 

Sin. 89° 39' 37^" : 46.8954 : : sin. 20' 22i" : BE. 
Log. tB 
Sin. 20' 22V 

Sin. 89° 39' 37" 
BE, in parts of a mile .27795 
In the triaDgle ABB, we have 



1.671130 

7.772824 
9.443954 
9.999992 
-1.443962 



Sin. 88° 47' 55 j < 
(See note) 



51.7861 : : sin. 31' 20" 
Log. 51.7861 
Sin. 31' 20" 



Sin. 88° 48' 
BD, .47210 
BR, .27795 
0.75005 
Log. 5280 feet in a mile, 
Height in feet, 3960 



Visible part, 
Invisible part, 
Whole mountain 



log. 



BD. 
1.714213 

7.959727 

9.673940 

9.999905 

-1.674035 

-1.875090 
3.722634 
3.597724 



2d Solution. 
We can also solve this problem by means of the triangle 
A CD alone. 

The angle CAD=90° 27' 5". (7--= 45', therefore the angle 
D=88° 47' 55". 
Sin. D : AG :: sin. GAD : CD. 



Log. 2 


.301030 


Cos. 10' 11.25" 


9.999998 




7.772822=sin. 20' 22.5". 


Log. 2 


.301030 


Cos. 20' 22.5" 


9.999993 




8.073845 =sm. 40' 45". 


Divide by cos. 40' 45" 


9.999971 




8.073874=tan. 40/45//. 



94 KEY TO [308 

Sin. 88° 47' 55" : AC : : sin. 90° 27' 5" : CD. 
AC. Bin. 90° 27' 5" 



CD. 



sin. 88° 47' 55" 



On AC ^^^^27' 5" 

ryn ^_^C(sin. 90°27'5"-sin. 88° 47' 55") /nx 

C1) - A0 - ^www ~ (1) 

But we have, Eq. (16), Plane Trigonometry, 

Sin. A— sin. B=2 cos. |(^+^)sin. i(A—B). 

Applying this to equation (1) above, and taking AC =3956 
miles, which is sufficiently near the truth for accuracy in the 
required result, the operation will be as follows : 



A= 


90° 27' 5" 
88° 47' 55" 
2)179° 15' 


log. 2 

COS. 

sin. 
=3956 


0.301030 




89° 37' 30" 


7.815909 


A-B= 


1° 39' 10" 




l(A-B) 


49' 35" 

Log. AC= 


8.159047 
3.597256 



-1.873242 
Sin. 88° 48' sub. 9.999905 



Log. of (CD-CA) in parts of a mile, -1.873337 
Feet in a mile 5280, log. 3.722634 

Log. (CD-CA) in feet, 3944.31 3.595971 

We will now compute the value of Am. From (Th. 18, 
Book III., Geometry,) we have 

2ACxAm= At 2 , nearly. 
Or, Am =§- G . 



308] PLANE TRIGONOMETRY. 95 

And taking again, -4(7=3956, we have 

Log. At 1.378742 

Log. 2 .301030 

AG 3.597256 



3.898286 3.898286 



-3.480456 
Log. 5280 3.722634 

Log. Am, 15.96 ft. 1.203090 
Hence, 3944.31 + 15.96=3960.27 feet =#2?. 

(20.) Note. — This problem is a singular and a very instructive one. "We can 
not directly make use of the given side or line 500 yards, and we are forced to 
make a similar geometrical figure by assuming another base, and using the 
given angles. 

The operation is as follows, 

Take^£=100. Make £^£=52° 12', 

GAD=41° 25'. ABC=47° 40', and 

GBD=48° 10'. 

These angles make the lines intersect 

at G and D, and CD must be computed, 

and if found to be 500, then 100 is the 

B true distance between the ships. After 

we have computed GD, we call it a. 

Then, because this figure is similar to the true figure, we 

have 

a : 500 : : 100 : AB. 

m A w 50000 

Whence, AB — . 

a 

The angle AGB=38° 43', and ADB=31° 58'. 

In the triangle ABG, we have 

Sin. 38° 43' : 100 : : sin. 93° 37' : BC (1) 




96 KEY TO . [308 

In the triangle ABB we have 

Sin. 31° 58' : 100 : : sin. 52° 12' : BB (2) 

Log. 100 sin. 93° 37' 11.999134 

Log. sin. 38° 43' 9.796206 

Log. BO, 159.561 " 2.202928 

Log. 100 sin. 52° 12' 11.897712 

Log. sin. 31° 58' 9.723805 

Log. BD, 149.247 2.173907 

CB+BB=308.808. CB -BB= 10.314. 



In the triangle OBB, we have 

308.808 : 10.314 : : tan. 65° 55' : tan. }(B-C). 

Log. 10.314 1.013427 

Log. tan. 65° 55 ! 10.349719 



11.363146 



Log. 308.808 2.489689 



Tan. ±(B-0) = 4° 16' 24" 8.873457 
i(Z>-f-C)=65°55' 
D =70° 11' 24" 
Sin. 70° 11' 24" : 159.561 : : sin. 48° 10' : CD. 

Log. 159.561 2.202928 
Log. sin. 48° 10' 9.872208 

12.075136 

Log. sin. 70° 11' 24" 9.973507 



Log. CI)=126M6 2.101629 ) 

Vsub. log. OB. 
Log. 50000 4.698970 \ 

Log. AB, 395.68 2.597341 



310] SPHERICAL TRIGONOMETRY. 97 

To obtain the true values of CB, BD, &c, we must mul- 
tiply the results in this computation by 3.9568. That is, the 
sides of the true figure are very nearly equal to 4 times the 
sides of this computed figure. 



SPHERICAL TRIGONOMETRY. 

(Page 310.) 

To solve right angled -spherical triangles, the student 
will find it most convenient to apply Napier's Circular 
points. 

Every triangle consists of three sides and two angles, be- 
sides the right angle, or five parts. 

Either one of these parts taken at pleasure may be called 
a middle part. Then there will be two adjacent parts, and 
two opposite parts. 

Then we can easily remember, that 

1st. Radius into the sine of the middle part, is equal to 
the product of the tangents of the adjacent parts. 

2d. Radius into the sine of the middle part, is equal to 
the product of the cosines of the opposite parts. 

The parts are the two sides, the complements of the hypo- 
tenuse, and the complement of the oblique angles. 

We can remember these rules by this consideration, that 
each rule expresses an equation. 

The first member of each of these equations is radius into 
some sine, the second member is the product of two tangents 
or two cosines. 

The student must rely on his own judgment, in selecting a 
part to be called the middle part. 



98 



KEY TO 



[356 



Let ABO represent any right 
angled spherical triangle, the 
right angle at B. b 

If the hypotenuse is greater than 90°, then the triangle 
will be represented by A' BO, and in that case one side will 
be greater than 90°, and the angle opposite to that side 
greater than 90°. 

For the sake of perspicuity, we recommend the natural 
construction of each triangle presented for solution. This 
course will banish all doubts from the mind of the student as 
to what the results should be, &c. The solution of one trian- 
gle will in all cases be the solution of a whole hemisphere of 
triangles, as we are about to show by the following exam- 
ples. 



RIGHT ANGLED SPHERICAL TRIGONOMETRY. 



(Page 356.) 

In a right angled spherical triangle 
ABO (right angles always at B), given 
^£=118° 21' 4", and the angle A = 
23° 40' 12". 

Kemark. — This triangle is repre- 
sented in the adjoining figure by the 
triangle A'BC, but we always operate on the triangle ABO 
of the figure, the one whose parts are each less than 90°. 

Whence, ^5=180°-118 o 21' 4" = 61° 38' 56"=c. 




SPHERICAL TRIGONOMETRY. 



99 



356] 

To this triangle we apply equation (16), page 335, Text 
book), observing that 

Sin. ^4'= cos. A and tan. &' = cot. b. 



Whence, 
Or. 



i?.cos. -4= tan. 61° 38' 56" cot. b. 



Cot. b = 



B.cos. A 



tan. c 
AC cot. 63° 42' 5" 

180° 
AG 116° 17' 55" 

For the angle c. 



19.961835 
10.267932 

9.693903 



As Sin. 63° 42' 5" : B 
B.sin. c= 
Sin. b 
0=79° 0' 34" 



sin. 61° 38' 56" : sin. O. 
19.944510 
9.952549 



9.991961 
Or, 100° 59' 26". Ans> 



(2.) In a spherical A ABO, given AB=53° 14' 20", and 
the angle ^ = 91° 25' 53". 

Here the proposed triangle is ABO, 
but we shall operate on its supplemental 
triangle A 1 BO. AB=A'B. The an- 
gle OA'B=90°-(1° 25' 53")=88° 34' 7". 
Eq. (16.) B. cos. 88° 34' 7"= tan. 
53° 14' 20".cot. A'C. 




Or, 



r» f #r/y - gjg: 1Q25 ' 53 " 
uot. ^0- Tan> 530^20" 



Cot. ^'(7 



88° 55' 51" 
180 



18.397585 

10.126658 

8.270927 



-4(7= 

For OB, we have 



91° 4' 9". ^?is. 
i2 : sin. A'O : : sin. A 1 : sin. .5'tf. 



100 

Whence, 



sin. JB'(7= 



KEY TO [357 

sin. 88° 55' 51", sin. 88° 34' 7". 



B 



Sin. 88° 55 } 51" 
Sin. 88° 34' 7" 
JB'<7, sin. 88° 12' 50" 
Sup. =BC, 91° 47' 10". Ans. 
Sin. A'C : sin. 90° : : sin. AB 
B.sin. ^J5 
Sin. ^(7=sin. J['(7 88° 55' 51" 
^ws. Sin. C=53° 15' 8" 

(3.) In this example, 



9.999925 
9.999864 

9.999789 



: sin. C. 

19.903707 
9.999925 
9.903782 




Let ^J5 = 102° 50' 25", and the angle 
BAC=US° 14' 37", but we shall operate on 
the supplemental triangle AB ] G) AB ] ~ 
77° 9' 35", and CAB' =66° 45' 23". 
Eq. (16), calling A the middle part, 
iZ.cos. 66° 45' 23"=cot. ACten. 77° 9' 35". 
B. cos. 66°, &c. 19.596202 

Cot. ^C=-r an AB 10.642190 

Ans. Cot. AC=cot 84- 51' 36" T954012 

For B ] C, we have 

iZ.cos. AC=co$. AB'.cos: B ] C: 
B.cos. AC, 84° 51' 36" 18.952258 

Cos. AB' 77° 9' 35" 9.346811 

Cos. B'C, 66° 13' 33" 9.605447 

And Supplement, BC, 113° 46' 27". Ans. 
For the angle C, we have 

Sin. AG : B : : sin. ^4£' : sin. C. 
Log. JR.sin. ^-S' 77° 9' 35" 19.989002 

Sin. AC, 84° 51' 36" 9.998250 

Sin. A CB>, sin. 78° 13' 4" 9.990752 

Supple. ACB=101° 46' 56". ^»s. 



357] 



SPHERICAL TRIGONOMETRY. 



101 




(4.) liGiAB'Cin the last cut represent 
the triangle to be solved, AB< =48° 24' 16", 
and^'C7, 59° 38' 27". 
Whence, 

i?.cos. A C= cos. 48° 24' 16".cos. 59° 38' 27". 
Cos. 48° 24' 16" 9.822082 

Cos. 59° 38' 27" 9.703652 



Cos. AC, cos. 70° 23' 42" 9.525734. Arts. 
For the angles, we have the following proportions. 
Sin. 70° 23' 42" : sin. 90° : : sin. 48° 24' 16" : sin. C 

: : sin. 59° 38' 27" : sin. A. 
Log. B.sin. 48° 24' 16" 19.873814 

Sin. 70° 23' 42" 9.974064 

Angle G. 52° 32' 56" 9.899750 



Log. iLsin. 59° 38' 27" 

sin. 70° 23' 42" 

Angle A, 66° 20' 40" 



19.935948 
9.974064 
9.961884. Ans. 



(5.) In this example, AB is 151° 23' 9", 
and BC ! , 16° 35' 14". 

We operate on the supplemental trian- 
gle A J BC. Whence 

A'B=28° 36' 51". And, 
B. cos. A'C> =cos. 28° 36' 51". cos. 16° 35' 14". 
Cos. 28° 36' 51" 9.943427 

Cos. 16° 35' 14" 9.981540 

£C\ cos. 32° 43' 9" 9.924967 

AC' 147° 16' 51". Ans. 




102 



KEY TO 



[357 



For the angle C, 

Bin. A'C : sin. 90° :: sin. A'B : sin. A OB. 

Log. iZ.sin. 28° 36' 51" 19.680253 

Sin. AC. 32° 43' 9" 9.732814 

sin. A OB, 62° 22' 35" 9.947439 

Its supplement, or AC'B=:117 37' 25" 

For the angle A, we have 

Asm. 16° 35' 14" 19.455568 

sin. 32° 43' 9" 9.732813 

Sin. ^=31° 52' 49" 9.722755. Ans. 




(6.) Here A'B, 73° 4' 31", ^'0>, 86° 12' 15". 
Kequired the other side and angles. 
Sin. A'O : sin. 90° : : sin. A'B : sin. BOA'. 
Log. B.sin. 73° 4' 31" 19.980771 
AC, sin. 86° 12' 15" 9.999046 
Sin. BOA', sin. 73° 29' 40" "9.981725. Arts. 
To obtain BO', we apply one of the equations in circular 
parts. 

K.cos. C7' = tan BO. cot, A'C 1 

Tan Ba - B ™*'° - 1.9.453484 

lan. BL - cot A}(ji 8 . 8 21819 

i?C"76 o 51'20" 10631665. Ans. 




(7.) Let AB'O' represent the proposed 
B' triangle. 

Then, ^J3'=47°26'35", and AO= 
118° 32' 12". 

We operate on the opposite supple- 
mental triangle, A' BO 1 . 



35?] SPHERICAL TRIGONOMETRY. 

Now, A'B=47° 26' 35". A'0' = 61° 27' 48". 

Sin. 61° 27' 48" : 22. :: sin. 47° 26' 35" : sin. O 

Rsin. 47° 26' 35" 19.867235 
Sin. C'=- 



103 



9.943748 
9.923487. ^rcs. 



Tan. BO. 



sin. 61° 27' 48" 
C'=^<7'£' = 56°58'44" 
Again, we have 

22.cos. 56° 58' 44" = cot. 61° 27' 48".tan. BO. 
Whence, 

.ff.cos. 56° 58' 44 " 19.736355 

cot. 61° 27' 48" 9.735427 

BO, tan. 45° 3' 40" 10.000928 

180° 

B'O 134° 56' 20". Ans. 
To find the angle A, we have 

Sin. A'O : sin. 90° : : sin. BO : sin. BA<0'. 

Log. B.sin. BO, 45° 3' 40" 19.849948 

Sin. A'O, 61° 27' 48" 9.943748 

Sin. BA'O, 53° 40' 58" 9.906200 

Supplement A, 126° 19' 2" Ans. 

(8.) Let A B'O represent the triangle, 
AB', 40° 18' 23", and AO, 100° 3' 7". Re- 
quired the other side and the oblique 
angles. 

We operate as before on the supple- 
mental triangle A' BO. AB'=A'B= 
40° 18' 23". ^'C"=79° 56' 53". 
1st. For the angle O, we have 

Sin. A'O : sin. 90° : : sin. A'B : sin. O. 
Log. R. sin. 40° 18' 23" 19.810821 

A'O, sin. 79° 56' 53" 9.993282 

O', sin. 41° 4' C" 0817539. Ans. 




104 KEY TO [357 

To find BO, take the angle O for a middle part, then by 
Napier's Circular Parts, we have 

Tan. BO, cot. A'C>=B.cos. 41° 4' 6". 
Log. i^.eos. 41° 4' 6" 19.877329 

^'tf', cot. 79° 56' 53" 9.248616 

BO, tan. 76° 46' 8" 10.628713 

Supplement BO, 103° 13' 52". ^s. 
For the angle A, we have 

Sin. ^4'(7 : sin. 90° : : sin. BO : sin. A'. 
Log. i2.sin. 76° 46' 8" 19.988316 

sin. 79° 56' 53" 9.993282 

A', sin. 81° 21' 7" 9.995034 

Supplement CM J5' = 98° 38' 53". Ans. 



(9.) In the right angled spherical triangle 
ABO, given AC, 61° 3' 22", and the angle A, 
49° 28' 12", to find the other parts. A 

As, Sin. 90° : sin. AG, 61° 3' 22" :: sin. ^,49° 28' 12" : sin. BO. 
AG, sin. 61° 3' 22" 9.942054 

A, sin. 49° 28' 12" 9.880852 

BO, sin. 41° 41' 32" 9.822906. Ans. 

For AB, we have the equation, 

i?.cos. AG— cos. ^4i?.cos. BO. 

B.cos. AG 19.684803 




Cos. AB. 



cos. BO 9.873162 



Cos. 49° 36' 6" 9.811641. Ans. 

For the angle O, we have 

Sin. AG, 61° 3' 22" : B. : : sin. ^JB, 49° 36' 6" : sin. (7. 
iZ.sin. AB 19.881703 

sin. AG 9.942054 



G, sin. 60° 29' 20" 9.939649. Ans. 



358] 



SPHERICAL TRIGONOMETRY. 



105 




(10.) Here we have given one side of a 
right angled spherical triangle, and its op- 
posite angle, to determine the other parts 
of the triangle, ABC. AB =29° 12' 50", 
and (7=37° 26' 21". 

In such cases the answers are said to be 
ambiguous, for the data give ns no indication of which tri- 
angle is intended, ABC or A ] BG. Because, AB'=A'B= 
29° 12' 50", and A'CB=ACB' = 37° 26' 21". 
To find AC, we have 
Sin. AB, 29° 12' 50" : sin. 37° 26' 21" 
iZ.sin. 29° 12' 50" 



Sin. AC-- 



sin. 37° 26' 21"" 
AC, sin. 53° 24' 13" 



; sin. AC : B. 
19.688483 
9.783846 



9.904637. Ans. 



Supplement A'C, 126° 35' 47". 
To find B'C. 

R.cos. AC=cos. AB' cos. B'C. 



Cos. B'C-. 



B.cos. 53° 24' 13" 



19.775372 
9.940916 



9.834456. Ans. 



cos. 29° 12' 50" 

B'C, cos. 46° 55' 2" 
Or its supplement. 

For the angle A, we have 

Sin. 53° 24' 13" : R. : : sin. 46° 55' 2" : sin. A. 

Log. iZ.sin. 46° 55' 2" 19.863542 

sin. 53° 24' 13" 9.904637 

Sin. A, 65° 27' 57" 

Or its supplement. 



9.958905. Ans. 



(11.) This example may require the solution of the tri- 
angle ABC, in adjoining figure, or of the triangle A' B'C. 



106 



KEY TO 



[358 

But the solution of either of these 
supplemental triangles is effected by the 
triangle A'BC, which is a common sup- 
plemental triangle. We have 
.4ifc^'i?' = 100 o 10'3". 
ACB=A'CB' ^90° 14' 20". 

Also, A'B=79° 49' 57", and A'CB=89° 45' 40". 

Sin. C : sin. A'B : : sin. 90° : sin. A'C 
Or, Sin. 89° 45' 40" : sin. 79° 49' 57" : : R. : sin. A'C. 
Whence, 

i^.sin. 79°49' 57" 




Sin. A'C 



19.993126 
9.999996 
9.993130 



sin. 89° 45' 40" 

A'C, sin. 79° 50' 8" 

180° 
AC, 100° 9' 52". Ans. 
By Eq. (20), Napier's Circular Parts, we have 
JS.cos. (7=cos. 79° 49' 57".sin. A. 
Cos.(7=cos.89 45'40"=:sin.0 14'20"=sin.860"=860sin.l". 
Log. sin. 1" 4.685575 (See page 288 Text-book.) 
Log. B. 860 12.934498 
Log. R. cos. C 17.620073 
Log. cos. 79° 49' 57" 9,246810 
BAC=sm. 1°21'12" 8.373263. Ans. 
For BC, we have 

B. : sin. A j C : : sin. A : sin. BC. 
Sin. A'C 79° 50' 8" 9.99313C 



Sin. A, 1° 21' 12" 
BC, sin. 1° 19' 55 u 



8.373263 
8.366393. Ans. 



358] 



SPHERICAL TRIGONOMETRY 



107 




(12.) This example may be represented 
by the triangle A' BO, or its supplemental 
triangle ABC, in the figure. 

In the triangle A'BC, we have 
Sin. C, 61° 2' 15" : sin. A'B, 54° 21' 35" : : 

[B. : sin. A'C. 



Sin. A'C= 



Rain. A'B 
sin. 61° 2' 15" 

A'C, 68° 15' 26" 
180° 



19.909925 
9.941976 
9.967949 



AC, 111° 44' 34". Arts. 
For BC, we have B.sin. BC= tan. ^'£. cot. (7. 
Tan. ^'jB 10.144485 

Cot. (7 9.743081 

Sin. 5(7, 50° 31' 32" 9.887566 
180° 
B'C= 129° 28' 28" 
For the angle BA'C, 

B.cos. BA'C=tsm. A'B.cot A'C. 
Tan. A'B 10.144485 

Cot, A'C 9.600770 

Cos. BAC=cos.BA'C 56° 12' 16" "9.745255 



180 c 



B'AC= 



123° 47' 44" 



(13.) This example may be made to cor- 
respond to the triangle ACB, or to the 
w triangle A'BC, because the opposite an- 
gles ACB, A'CB' are equal, and AB= 
A'B. 

But we shall operate on the triangle 
A'BC, which is supplemental to each of the other two. 




108 key to [35§ 

Because AB = 121° 26' 25", A'B=58° 33' 35". 
Because ACB =111° 14' 37", A'CB=68° 45' 23". 

For the side A ! C, we have 

Sin. C : sin. ^'5 : : B. : sin. -4'tf. 
Log. B.sin. 58° 33' 35" 19.931043 

Sin. C, 68° 45' 23" 9.969439 

A'C, sin. 66° 15' 38" 9.961604 

Supplement, or AC, 113° 44' 22". Arts. 

For the angle A, or BAG, we have 

Tan. ^'£.cot. A'C=B.cos. A J =B.cos. A. 
Or, Cos. A = tan. 58° 33' 35".cot. 66° 15' 38". 

Tan. 58° 33' 35" 10.213698 

Cot. 66° 15 ; 38" 9.643246 

A, cos. 43° 59' 55 n 9.856944 
Or, supplement, B'A'C, 136° 0' 5". Ans. 

For the side BC, we have 

B. : sin. u4'(7 : : sin. A' : sin. BO. 

Sin. ^'(7 9.961604 

Sin. A' 9.841760 



BC, sin. 39° 29' 3" 9.803364 

Supplement, or B ] C, 140° 30' 57" AC. 



QUADRANTAL TRIANGLES. 

Quadrantal spherical triangles have one side equal to 90 
degrees, and all such triangles can be solved by right angled 
spherical trigonometry, as illustrated in the Text-book. 

The following are not solved in the Text-book. 



361] 



SPHERICAL TRIGONOMETRY 



109 



PRACTICAL PROBLEMS. 



(Page 361.) 




(1.) In a quadrantal triangle, given the quadrantal side? 
90°, a side adjacent, 67° 3', and the included angle, 49° 18' 
to find the other parts. 

r The remaining side is 53° 5' 44" ; the angle oppo- 
Ans. < site the quadrantal side, 108° 32' 29" ; and the re- 
C maining angle, 60° 48' 54". 

The triangle corresponding to this ex- 
ample is represented by A P C. AP =90°, 
P(7=67° 3', and the angle at P= 
49° 18'. 

We operate upon the triangle ABC, 
having AB=P=49° 18', and BG= 
90°-(67 o 3')=22°57'. 

For AC, 

B.cos. AC=cos. 49° 18'. cos. 22° 57' 
Cos. 49° 18' 9.814313 

Cos. 22° 57' 9.964187 

AC, cos. 53° 5' 44" 9.778500. Ana. 

For the angles A and C, we have the proportions 
Sin. AC : B. : : sin. AB : sin. C 
: : sin. BC : sin. A. 
For C For A. 

Log. iZ.sin. 49° 18' 19.879746 Log. iZ.sin. 22° 57' 19.590984 
sin. 53° 5' 44" 9.902894 sin. AC 9.902894 



C, 71° 27' 31" 9.976852 A, 29° 11' 6" 9.688090 

Sup. ACP, 108° 32' 29". Ana. Com. PAC, 60° 48' 54" Ans. 



110 



KEY TO 



[361 

(2.) In a quadrantal triangle, given the quadrantal side, 
90°, one angle adjacent, 118° 40' 36", and. the side opposite 
this last-mentioned angle, 113° 2' 28", to find the other 
parts. 

/ The remaining side is 54° 38' 57" ; the angle oppo- 
Ans. I site, 51° 2' 35" ; and the angle opposite the qaud- 

( rantal side, 72° 26' 21". 
The triangle ^4'PCcorresponds to this 
example. A'PC=A'B=118° 40' 36". 

Whence, AB=61° 19' 24", A'C= 
113° 2' 28" ; therefore AG=66° 57' 32". 
BC is the complement of PC, the side 
required. 

For BO in the triangle ABC, we have 

jff.cos. AC=cos. AB. cos. BC. 

P.cos. 66° 57' 32" 19.592611 

9.681120 




Cos. BC-- 



cos. 61° 19' 24" 

BC, cos. 35° 21' 3" 
Whence, P(7=54° 38' 57". Ans. 

For the angles A and C 



9.911491 



Sin. AC : B. : : sin. 


AB : 


sin. C 


Sin. AC : B. : : sin. 


BC : 


sin. A. 

For C. 


Log. P.sin. 61° 19' 24" 




19.943168 


Sin. AC, m° 57' 32" 




9.963894 


Sin. (7=72° 26' 21" 




9.979274. 
For A. 


Log. P.sin. 35° 21' 3" 




19.762365 


Sin. AC, 66° 57' 32" 




9.963894 


Sin. 38° 57' 26" 


9.798471 


Com. 51° 2' 34' 


. Ans 


. PA'C 



Ans. 



tfiirf] 



SPHERICAL TRIGONOMETRY. 



Ill 



(3.) In a quadrantal triangle given the qnadrantal side 
90°, and the two adjacent angles, one 69° 13' 46", the other 
72° 12' 4", to find the other parts. 

r One of the remaining sides is 70° 8' 39". the other 
Ans. < is 73° 17' 29", and the angle opposite the quad- 
t rantal side is 96° 13' 23". 

The triangle is represented by PAC. 
Taking the angle P^-69° 13' 46", then 
AB=69° 13' 46", and the angle BAG 
must equal 90°-(72° 12' 4"), or, 
17° 47' 56"= A. 

Now, by taking the angle A for a 
middle part, we have the equation, 

Cot. AGMn. AB^B.cos. A. 




Whence, 



Cot. AC-- 



B.cos.A,17°47 , 56 u 
'' tan. 69° 13' 46" 

AC cot. 70° 8' 39" 



19.978698 
10.421044 



9.557654. Ans. 
For the side BC, of the triangle ABO, we have 
B. : sin. AC : : sin. A : sin. BC. 
Sin. AC 70° 8' 39" 9.973382 



Sin. -4, 17° 47' 56" 



9.485262 



BC, sin. 16° 42' 31" 9.458644 

Complement, PC, 73° 17' 29". Ans. 

For the angle C, of the triangle ABC, we nave 
j&eos. A C= cot. A. cot. (7. 
2?.cos. ^(7 19.531037 

Cot. A 10.493436 

Cot. C, 83° 46' 37" 9.037601 

Supplement, PC A =96° 13' 23". Ans. 



112 



KEY TO 



[362 



(4.) In a quadrantal triangle, given the quadrantal side, 
90°, one adjacent side, 86° 14' 40", and the angle opposite to 
that side, 37° 12' 20", to find the other parts. 

r The remaining side is 4° 43' 2" ; the angle opposite 
Ans. < 2° 51' 23" ; and the angle opposite the quadrantal 
C side, 142° 42' 3". 



The triangle is represented by PAO, 
PO=86° 14' 40", and PAC=37°12'20 n - 

Whence, the angle A, of the triangle A 
ABC=52° 47' 40", and BO, 3° 45' 20". 
For AC, we have 

Sin. A : sin. BC : : R. : sin. AC. 
Log. iZ.sin. BC, 3° 45' 20" 
sin. A, 52° 47' 40" 
Sin. A a 4° 43' 2" 




p' 

18.816240 
9.901170 
8.915070. Ans. 



For AB, we have 

iZ.cos. A — cot. -4(7. tan. ^4^. 
iZ.cos. 52° 47' 40" 19.781523 

Cot. 4° 43' 2" 11.083454 

Tan. AB, 2° 51' 23" 



For the angle C, of the triangle ABC, 
B.cos. AC— cot. .i.cot. C. 
B.cos. 4° 43' 2" 
cot. 52° 47' 40" 
Cot. C, 37° 17' 57" 



8.698069. Ans. 



19.998527 
9.880353 



10.118174 
Supplement, PC A =142° 42' 3". .4 ws. 



(5.) In a quadrantal triangle, given the quadrantal 
side, 90°, and the other two sides, one 118° 32' 16", the 



363] 



SPHERICAL TRIGONOMETRY, 



113 




other 67° 48' 40", to find the other parts— the three 
angles. 

r The angles are 64° 32' 21", 121° 3' 40", and 77° 11' 
Ans. < 6" ; the greater angle opposite the greater side, 
( of course. 

This problem requires the solution 
of the triangle A' PC, or P ] AC, each 
, one may by hypothesis correspond with 
the data. That is, A'Oot P'(7=118° 
32' 16", and PC or AC=6T 48' 40". 
We will take P'AC. Then, BC= 
118° 32' 16", less 90°. 
That is, JBC=28° 32' 16". 

In the triangle ABC, we have AC and BC. 
To find the angle A, or CAB, we have 
Sin. AC, 67° 48' 40" : B. : : sin. 28° 32' 16" : sin. A. 
Log. iZ.sin. 28° 32' 16" 19.679191 

sin. 67° 48' 40" 9.966585 

Sin. A, 31° 3' 40" 9.712606 

Add 90° and P'AC=121° 3' 40". Ans: 
The side AB, is the measure of the angles P and P', and 
the angle C is also an angle in the triangle P'A C. 
For AB, B. : cos. AB : : cos. BC : cos. AC. 



Cos. AB= 



B.cos.AC, 67° 48' 40" 



cos. BC, 28° 32' 16" 
Angle P'=AB, 64° 32' 21" 
For the angle C, we have 

Sin. AC : B. : : sin. AB 
Log. B.sin. AB, 64° 32' 21" 
Log. B.sin. AC, 67° 48' 40" 
sin. C, 77° 11' 6" 



19.577102 




9.943743 




9.633359. 


Ans. 


sin. C 




19.955630 




9.966585 




9^989045. 


Ans. 



114 



KEY TO 



[3©^ 



(6.) In a quadrantal triangle, given the quadrantal side, 
90°, the angle opposite, 104° 41' 17", and one adjacent side, 
73° 21' 6", to find the other parts. 

C Kemaining side, 49° 42' 16" ; remaining an- 
nS ' \ gles, 47° 32' 38", and 67° 56' 13". 

This example refers to the triangle 
APC, because the given angle is greater 
than 90°. We must operate on the 
triangle ABC 

The angle A CB = 180- (104° 41' 17") 
=75° 18' 43", and AG, 73° 21' 6". 
For AB } we have 

R. : sin. AC : : sin. C : sin. AB. 
Whence, Sin. AB — sin. ^Csin. C 

Sin. 75° 18' 43" 9.985571 

AC, sin. 73° 21' 6" 9.981402 

APC=AB, sin. AB, 67° 56' 13" 9.966973. Ans. 

For the side BC, we have 

B. : cos. AB : : cos. BC : cos. ^4(7. 

R.cw.AC 19.457120 

9.574757 




Cos. BC= 



cos. AB 
Cos. 5(7,40° 17' 44" 



8.882363 
Complement, P<7=49° 42' 16". Ans. 

For the angle A, we have 

Sin. AC : R. : : sin. 5(7 : sin. ^. 
Log. iZ.sin. BC 19.810723 

sin. 73° 21' 6" 9.981402 

Sin. A, 42° 27' 22" &829321 

Complement, PA (7=47° 32' 38". Ans. 



8©7] SPHERICAL TRIGONOMETRY. H5 

OBLIQUE ANGLED SPHERICAL TRIGONOMETRY. 

PRACTICAL EXAMPLES. 

(Page 367.) 

Note. — Here, as in Plane Trigonometry, the sides are re- 
presented by a, b, c, and the angles opposite by A, B, C, 
that is, A opposite a, B opposite b, and opposite c. 

(1.) Given 5 = 118° 2' 14", 0=120° 18' 33", and the in- 
cluded angle .^=27° 22' 34", to find the other parts. 

Here we have two sides, and the in- 
cluded angle. 

This triangle is represented by ABO, 
but we operate on the supplemental tri- 
angle A'BO. 

We may let fall the perpendicular CD, 
dividing the triangle A J B into the two right angled spher- 
ical triangles, A'DC, and BBC. 

Or we may solve either triangle ABC or A } B C directly, 
by applying Equations (8) and (9), (page 350, Geometry), 
which are 

Tan. x( g+jB ) = cot -^.cos. i(c-&) 

' v ' cos. i{c + b) v ' 

Tan. i{C _ B)=: ^.iA. S m. i (c-b) 

- v J sin. \(c + b) K ' 

c=120° 18' 33". Whence, ^'£=59° 41' 27". 

b =118° 2' 14" 




Sum 238° 20' 47" half sum 119° 10' 23.5". 
Diff. 2° 16' 19" half dim 1° 8' 9.5". 



116 KEY T ° [368 

Cot. f4 = 13° 41' 17" 10.613406 10.613406 

cos. 1° 8' 9.5" 9.999915 sin. 8.297218 

20.613321 1^910624 

Cos. 119° 10' 23.5" 9.687932?* sin. 9.941090 

Tan. ±(C+B)(83° 13' 42") 10.925389^ 
Tan. \{C-B){ 5° 19' 34") 8.969534 

Note. — The cosine of an arc greater than 90° is negative. 
Hence the cosine 119° is minus, and we place n against the 
log. to show that it is negative. And since tan. \(C+B) is 
negative also, the arc mu§t terminate in the 2d quadrant * it 
is therefore the supplement of 83° 13' 42" ; hence 
i(G+B) 96° 46' 18" 
\{C-B) 5° 19' 34" 

Sum C, 102° 5' 52" ) 
Diff. B, 91° 26' 44" \ 
For the side BO, we subtract G from 180°, giving 77° 54' 8" 
for the angle A'CB. 
Sin. 77° 54' 8" : sin. 59° 41' 27" : : sin. 27° 22' 34" : sin. a. 
Sin. 59° 41' 27" 9.936170 

Sin. 27° 22' 34" . 9.662597 

19.598767 
Sin. 77° 54' 8" 9.990246 



Sin. a, 23° 57' 13" 9.608521. Ans. 

(2.) Given 

.4=81° 38' 17" 7 

B=70° 9' 38" } ^° ^ n( ^ a > b) an ^ c - 

(7=64° 46' 32" J By formula (W), (page 348, Geom.) 

2)216° 34' 27" . , /-cos. S. cos. (8-A)\t 

Sin. -k(X I ; =r — ; -^ I . 

S 108° 17 ; 13.5" v sm - -^.sin. C I 

81° 38' 17" 



S-A 26° 38' 56.5" 



368] 



SPHERICAL TRIGONOMETRY. 



117 



Note. — The arc S being greater than 90°, its cosine is 
minus } and subtracting a minus quantity as the sign in the 
formula indicates, makes it plus. 



Cos. &, 108°17' 13.5" 

Cos.(S-A), 26° 38' 56.5" 

Sin. B, 70° 9' 38" 9.973427 ) 
Sin. 0, 64° 46' 32" 9.956479 \ sum, 

Sin. |a.(Kadius unity) = 
For the radius of tables, add 
Tabular sin. of ia=35°2' 6.3" 

2 



9.496623 

9.951226 

19.447849 

19.929906 

2)-1.517943 

-1.758971 

10. 

9.758971 



Because, 



a=70°413". Ans. 
sin. b sin. a 
sin. B~ sin. A' 



Therefore, Sin. b = sin. B. JEL5. 

sin. A 

Sin. a. 70° 4' 13" 9.973179 

Sin. A, 81° 38' 17" 9.995358 

Sin. B 
b } 63° 21' 24" 



Sin. c=sin. C— 



sin. a 



sin. ^4 



-1.977821 
9.973427 
9.951248 



sin. C 



-1.977821 

9.956479 



c=59°16'21" 9.934300 

Ans. 



(3.) Given a, 93° 27' 34", b, 100° 4' 26", and c, 96° 14' 50", 
the three sides to find the three angles. 

By formula, on page 343. 

Cos. jA=( sla - 8 - f^-^Y^dhxs unity.) 
\ sin. b. sin. c. / 



118 



KEY TO 

a 93° 27' 34" 

b 100° 4' 26" sin. 9.993252 

c 96° 14' 50" sin. 9.997413 



[368 



2)289° 46' 50" 19.990665 sum, 



# 144° 53' 25" sin. 
93° 27' 34" 



(S-a) 51° 25' 51" sin. 
Sub. sum as above, 

Cos. \A } 47° 19' 32" 

2 



9.759777 



9.893127 

19.652904 

19.990665 

2)-1.662239 

9.831120 adding 10. 



.4, 94° 39' 4" 



We may now obtain the other angles by Equations (8) 
and (9), (page 350, Geom.) 
5 = 100° 4' 26" 
c= 96° 14' 50" 



±(b + c) = 98° 9' 38" 
K&-c)= 1°54'48" 

Cot. i A, 9.964707 9.964707 

Cos. i(Z>-c) 9.999758 sin. 8.523587 

19.964465 18.488294 

Cos. i(b+c) 9.152128ft sin. 9.995580 

Tan. i(B + G) 10.812337ft tan. \(B-0) 8.492714 
i(B + G)= 98° 45' 27" 
^B-C)= 1° 46' 52" 
£=100° 32' 19" 
C= 96° 58' 35" 

(4.) Given two sides, b, 84° 16', c, 81° 12', and the angle 

O, 80° 28', to find the other parts. 




368] SPHERICAL TRIGONOMETRY. H9 

When this triangle is constructed, we find that the data 
will correspond equally well to the tri- 
angle ABC, and to AB ] C, in the adjoin- 
ing cut. 

- Hence the result is said to be ambig- B ' 
uous. In sucli cases the operator is 
expected to determine both results. 

Observe that AB=AB', hence the triangle ABB 1 is 
isosceles, and AD the perpendicular from A bisects BB' in 
D, making two right angled spherical triangles, ABC and 
ABB 1 . Their sum is the triangle ACB', and their differ- 
ence ABC. 

For the angles at B, or at B\ we have the proportion 

Sin. c, 81° 12' : sin. C, 80° 28' : : sin. b, 84° 16' : sin. B. 

a . D sin. 84° 16'.sin. 80° 28' 

bm. Jb— : ■ . 

sin. 81° 12' 

Sin. 84° 16' 9.997822 

Sin. 80° 28' 9.993960 

19.991782 
Sin. 81° 12' 9.994857 

Sin. B\ or ABB, 83° 11' 24" 9.996925 

Supplement, 96° 48' 36" =ABC Ans. 

By Equation (16), Napier's Circular Parts (page 335, 
Geometry), we have 

Tan. <7Z>.cot. b=B.cos. C=B.cos. 80° 28'. 

iZ.cos. 80° 28' 19.219116 



Tan. CD= 



cot. 84° 16' 9.001738 



CD, tan. 58° 46' 31" 10.217378 

In like manner, we obtain BD, from the triangle ABD. 



120 KEY TO 


[368 


mi ,- m -or, i?.C0S. 83° 11' 24" 

That is, Tan. I?D_ cot c> 81 o 12 , 


19.074002 
9.189794 


BD, tan. 37° 27' 2" 


9.884208 


But (71?, 58° 46' 31" 




Sum G#, or a, 96° 13' 33". 


Ans. 


Diff. CB, or a, 21° 19' 29". 


Ans. 


For the angles A, we have 




Sin. m sin. (7 4 

__. =-; , or sm. m— sm. 

bin. c& sm. c 


sin. C 

a.~ 

sm. c 


Sin. a. 21° 19' 29" 9.560688 sin. a. 96 


°13'33" 9.997431 


Sin. C- sin. c -1.999103 


-1.999103 


u4, sin. 21° 16' 43" 9.559791 A, sin. 97 


° 13' 45" 9.996534 



Ans. 
(5.) Given one side c, 64° 26', and the adjacent angles 
B A, 49°, and B, 52°, to find the other 
parts. 

Let misrepresent the triangle, and 
from one extremity of the given side, 
c D let fall the perpendicular BD, mak- 
ing the two right angled spherical triangles, ADB and GDB. 
By Equation (16), Napier's Circular Parts (Geom., p. 335), 
we have Tan. AD. cot. 64° 26'=JB.cos. 49°. 

B.cos. 49° 19.816943 

Cot. 64° 26' 9.679795 




Tan. AD, 53° 54' 10.137148 
For the angle ABD, we have 

Sin. 64° 26' : B. : : sin. 53° 54' : sin. ABD. 

B.an. 53° 54' 19.907406 

Sin. 64° 26' 9.955247 

Angle ABD = 63° 35' 51" 9.952159 
ABC 52 



Angle (7£D=11°35'51" 



3©8] SPHERICAL TRIGONOMETRY 121 

In the triangle ABD, we have 

B. : sin. 64° 26' : : sin. A, 49° : sin. BD. 
Sin. 49° 9.877780 

Sin. 64° 26' 9.955247 

BD, sin. 42° 54' 26" 9.833027 

For BO, or a of the triangle ABC, we use the equation 
Tan. BR cot. BC=B:cos. 11° 35' 51". 
B.cos. 11° 35' 51" 19.991042 

Tan. 42° 54' 26" 9.968246 



Cot. BO, 43° 29' 49" 10.022796. Ans. 

For A O, or h, we have 

Sin. 49° : sin. 43° 29' 49" : : sin. 52° : sin. AC. 
Sin. 43° 29' 49" 9.837788 

Sin. 52° 9.896532 

19.734320 
Sin. 49° 9.877780 

I, or AC, sin. 45° 56' 46" 9.856540. Ans. 

For the angle C, we have 

B.cos. BC= cot. CBD. cot. BOD. 
B.cos. 43° 29' 49" 19.860584 

Cot. CBD, 11° 35' 51" 10.687769 

Cot. BCD, 81° 31' 56" 9.172815 

BCA=98° 28' 4" ^«w. 

(6.) Kesult obvious. 

(7.) Given two sides and an angle opposite one of them 
to determine the other parts. 

a=77 25'll",c = 128° 13' 47", and the angle 0=131° IV 12". 
To find the angle A, we have 

Sin. A _sin. O sin. 131° 11' 12" _ cos. 41° 11' 12" 
Sin. a "sin. c ~"sin. 128° 13' 47" "cos. 38° 13' 47" 



122 



KEY TO 



[368 



Whence, Sin. A 



sin. a. 



COS. 


41° 


11' 12" 


COS 


38° 


13' 47" 


COS. 


38° 


13' 47" 



9.989446 sin. a 
9.876546 



19.865992 
9.895166 



Sin. A, 69° 13' 59" 9.970826. Ans. 

We have now to determine B and b, and the process will 
be apparent after Ave construct the triangle, as represented 
in the adjoining figure. 

The data gives the triangle ABC ; ABC 
is a supplemental triangle. From B, let 
fall the perpendicular BD. 

In the right angled triangle A B'D, 
we have the angle A, and the hypo- 
c tenuse AB ] . AB being the supple- 

ment of c. 

From the two triangles ABB and C'DB' we can obtain 

AB, and DC, and their sum taken from 180°, will give 

AC, or b. . 

The angle DOB, is the supplement of C, which is 

48° 48' 48" 
By Napier's Circular Parts, we have 

Cot. A B. tan. AD=fi.cos. A. 




Or, Tan. AD 



iE.cos. 69° 13' 59" 
cot, 51° 46' 13" 

Tan. AD, 24° 13' 56 u 

, ™ ^ ^.cos. 48° 48' 48" 
Also, Tan. ^= cot .^ 77 o 25 ' 11" 

Tan. G*D, 71° 16' 44" 
AD + Z><7 =95 3 30' 40" 
Supplement -4(7, 84° 29' 20'' = 6. 



19.549698 
9.896396 
9.653302 

19.818566 

9.348626 

1O469940 



^4^5. 



368] SPHERICAL TRIGONOMETRY. 123 

Lastly, for the angle B, we have 

Sin. B sin. A . ^ sin. b. sin. A 

-^ — =-= , or sua. B—~ = . 

Sin. b sin. a sin. a 

Sin. b 9.997988 

Sin. A 9.970826 



19.968814 
Sin. a 9.989446 



Sin. B, 72° 28' 42" 9.979368. Ans. 



(8.) Given a= 68° 34' 13" to find A, B, and 0. 

b= 59° 21' 18" sin.cona. 065328 
c = 112° 16' 32" sin. com. 033684 



2)240° 12' 3' 



Use formula (T), 120 3 6' 1.5" S. sin; 9.937090 

(p. 343, Geom.) 68° 34' 13" 

51° 31' 48.5" (S-a) sin. 9.893726 
2)19.929828 

Cos. 22° 43' 19" 9.964914 

2 



A, 45° 26' 38". Ans. 

Now, Sin. B— sin. &.— — - — ■, and sin. (7= sin. c. -r— : — . 
sm. a sin. a 

Sin. 5= 9.934672 sin. c = 9.966316 

Sin. ,4— sin. a= -1.883936 -1.883936 



B, sin. 41° IV 30" 9.818608 C, 134° 53' 55" 9.850252 



Note. — "We take C greater than 90°, because c was given greater than 
90°. The logarithm gives 45° 6' 5", and its supplement is the angle required. 



124 key to [369 

(9.) Same formula as applied to the preceding. 

Given a= 89° 21' 37" to find A, B, and C. 



b= 97° 18' 39" sin. 
c= 86° 53' 46" sin. 


j., ±j 

com 
com. 

sin 


003546 
000637 


2)273° 34' 2" 
136° 47' 1" S. sin. 
89° 21' 37" 


9.835536 


47° 25' 24" (S-a) 

Cos. \A, 44° 28' 40" 

2 


9.867097 

2)19.706816 

9.853408 


A, 88° 57' 20". Ans 
Now we have, 





a . D . , sin. A y • n - sm - -^ 

Sm. B—<&\vl. o. , and sm. G r =sin. c. . 

sm. a sm. a 

Sin. 5 9.996454 sin. c 9.999362 

Sin. .4 -sin. a -1.999955 -1.999955 

B, sin. 97° 21' 26" 9.996409 (7=88° 47" 17" 9.999317 



(10.) Given a=31° 26' 41", c=43° 22' 13", and the angle 
^4=12° 16', to find the other parts. 

^ This example applies to the adja- 

cent figure. 

ABC and ABO, either one, will 
D correspond with so much of the 

data as is given. Hence the result is ambiguous. 

But, Sin. C— sin. c. -.— — . 

sm. a 




369] SPHERICAL TRIGONOMETRY. 125 



c, sin. 43° 22' 13" 




9.836774 


A, sin. 12° 16' 




9.327281 




19.164055 


a, sin. 31° 26' 41" 




9.717400 


O, sin. 16° 14' 27" 




9.446655 


C, 163° 45' 33". 


Ans. 





Or, 

In the right angled spherical triangle ADJB, we have 

B. : sin. c : : sin. A. : sin. BD. 
Sum as ahove, omitting radius. 

Sin. BD, 8° 23' 22" =9.164055. 
In the same triangle we have, 

iZ.cos. c=cos. AD.cos. BD. 



B.C08. c 

Cos. BD 




19.861493 
9.995327 


Cos. ^D=42 c 


> 42' 37" 


9.866166 


right angled triangle CBD, we have 
jR.cos. a=cos. CD.cos. BD. 
iZ.cos. a 19.931022 
Cos. BD 9.995327 


Cos. CD, 


30° 24' 57" 


9.935695 



AD= 42° 42' 37" 

&um= AC 1 =b, 73° 7' 34". ^s.. 

Diff.=^(7=&, 12° 17' 40". ^ws. 

To find the angles at B, we have the following propor- 
tion. 

Sin. a : sin. A : : sin. 12° 17' 40" : sin. ABC. 
Sin. A 9.327281 

Sin. 12° 17' 40" 9.328248 

18.655529 
Sin. a 9.717400 

ABC, 4° 58 f 30" 8.938129 



126 



KEY TO 



[369 



Sin. a. : sin. A. : : sin. 73° 7' 34" : sin. ABC. 



Sin. A 

Sin. 73° 7' 34" 

Sin. a 

C'BA, 22° 56' 16" 

180° 



9.327281 

9.980888 

19.308169 

9.717400 



9.590769 



ABO 157° 3' 44" 




(11.) In a triangle ABC, we have given, ^ 56° 18' 40", 
.£, 39° 10' 38", and AD, 32° 54' 16", a segment of the base 
made by a perpendicular let fall from the angle C, on to the 
side AB, to determine the triangle. The angle C being 
obtuse. 

This is an ambiguous case. 
For, in the lune B" B, let.fl"= 
J5'=39° 10' 38". Take a point, 
C, nearer to B" than to B', and 
draw CB equal to CB". Then the triangle CB'B will be 
isosceles, and we shall have CBB" = B"=B'. Suj^pose CA, 
and the perpendicular CD, be drawn, making CAB— 56° 18 
40", and ^Z>=32° 54' 16". Now since B' = CBA, the given 
parts belong equally to the two triangles, CAB, and CAB'. 
It will be observed, however, that CB and CB' are supple- 
ments of each other, because CB — CB". 

Tan. 32° 54' 16". cot. AC=B. cos. A. 56° 18' 40". 

19.744045 



i^.cos. 56° 18' 40" 

Tan. 32° 54' 16" 

(5) cot. AC, 49° 23' 41" 



9.810931 
9.933114. A ns. 



H69] SPHERICAL TRIGONOMETRY. 

For CD, we have 

B. : sin. AC : : sin. A 
Sin. AC 
Sin. ^ 
Sin. CD, 39° 10' 35" 



127 



sin. CD. 

9.880363 
9.920155 



9.800518 
Now in the right angled triangle CDB, we have 

Sin. B : sin. CD : : B. : sin. CB. 
B.sin. CD 19.800518 

Sin. B 
Sin. CB, 89° 40' 



9.800525 
9.999993 



Or, 



CB' =90° 20' 



Note. — It will be observed that arcs which differ from 90° by less than 1°, 
can not be determined accurately to seconds, when the sine is used ; and we 
can not use the cosine or tangent in this case. 

For the angle A CD, we have 

Sin. AC : B, : : sin. AD : sin. ACD. 

B.sm.AD, 32° 54' 16" 19.734991 

Sin. AC, 49° 23' 41" 9.880363 

Sin. ACD, 45° 41' 12" 9.854628 

To determine the angles, DCB and DCB ', we have 
B.cos. DCB =tan. CD.cot. CB. 
B.cos. DCB'=tan. CD.cot. CB'. 



Tan. OZ)=9.911102 

Cot. CB =7.764761 



Tan. CD =9.911102 

Cot, CB =7.764761^ 



Cos. DCB = 7.675863 
DCB= 89°44' 
Add ACD= 45° 41' 



Cos. DCB' = 7.675863^ 
DCB'= 90° 16' 
ACD= 45° 41' 



A CB =135° 25' Ans. 



ACB' = 135° 57 ^ 



ns. 



128 



KEY TO 



[369 



For the sides AB and AB' we have 



Sin. B : sin. AG 
Sin. B : sin. AG 
Sin. ,4(7= 9.880363 
Sin. ^<7£ = 9.846304 



Sin. 5: 



19.726667 
9.800525 



Sin. AB = 9.926142 
AB= 122° 29' 



sin. ACB : sin. AB. 
sin. ^0#' : sin. AB. 
Sin. -4(7= 9.88,0363 
Sin. ,4GB'= 9.842163 



Sin. 5' 



19.722526 

9.800525 



Arts. 



Sin. ^£'= 9.922001 

u4J?'= 123° 19' -4ws. 



(12.) Given the angles A, B, C, and required the sides 
a, b, c. (Prop. 6, Sec. L, Spherical Geometry.) 



^=80° 10' 10" 
£=58° 48' 36" 
(7= 91° 52' 42" 



sup. 99° 49' 50" 
121° 11' 24" 

88° 7' 18" 



sin. com. 
sin. com. 



.067803 
0.000233 



2)309° 8' 32" 


sin. 

sin. 




S, 154° 34' 16" 

a, 99° 49' 50" 


9.632852 


(S-a) 54° 44' 26" 


9.911981 


Cos. J A, 50° 10' 49" 

2 


2)19.612869 
9.806434 



100° 11' 38" 
Supplement a =79° 38' 22". Arts. 

Note. — The preceding process strictly corresponds to theory, but the result 
will be the same, if we take out the arc, whose sine corresponds to the given 
logarithm, and the double of that arc will be the side a. 

Thus, the sine of 39° 49' 11", is 9.806434, and the double of 39° 49' ll", is 
T9° 38' 22", or a. 



Now 



, bin. 6 = sin. B. -. - A 

sm. A 



, . • n sm. a 

, and sm. c = sin. U. -. 7 

sm. A 



373] SPHERICAL TRIGONOMETRY. 129 

Sin. B, 9.932197 sin. C, 9.999767 

Sin. a -sin. A. -1.999284 -1.999284 



5=58° 39' 16" 9.931481 c=86° 12' 50" 9.999051 



SECTION V. 

APPLICATION OF SPHERICAL TRIGONOMETRY TO THE 
SOLUTION OF PROBLEMS IN ASTRONOMY. 



(Page 373.) 



(2.) In latitude 42° 40' N. ; when the sun's declination is 
23° 12' K, what time will the sun set ? 

Tan. 42° 40' 9.964588 

Tan. 23° 12' 9.632053 



Sin. 23° 16' 7" 9.596641 

This arc, 23° 16' 7" ; reduced to time at the rate of 4 min- 
utes to one degree, gives P 33 w 4 5 . 
Add 6 



Sun sets, 7 h 33 m 4 s . sun rises. 4 h 26 m 56 s , 



(3.) What time will the sun set in latitude 42° 4' north, 
and sun's declination 15° 21' south ? 

Tan. 42° 40' 9.964588 

Tan. 15° 21' 9.438554 



Sin. 14° 39' 21" 9.403142 



130 



KEY TO 



[373 



This arc corresponds to 58™ 37 s in time, which is the inter- 
val between six o'clock and sun set, and as the observer is 
north, and the declination south, the sun must set before six, 
that is, at 5 h V n 23 s , apparent time. The sun must rise 
same day at 6 h 58™ 37% A.M. 



(4.) Lat. 52° 30' north (London), and the sun's declination 
18° 41' south. Kequired, times of sunrise and sunset. 
Tan. 52° 30' 10.115019 

Tan. 18° 42' 9.529535 

Sin. 26° 10' 30" 9.644554 

This arc corresponds in time to 1* 44™ 42 ? , to which add 
6 h for sunrise, and subtract it from six hours for sunset. 

Kises,UM4-42M 

Sets. )4Ai*»i«^ apparenttime ' 



Whence, 



Ans. 



(5.) This problem is clearly represented by the adjoining 
cut. NS is the earth's axis, NFS the meridian, EQ the 
equator, en the parallel of declination. 

The right angled spherical triangle 
amn, will give the position of the 
sun, and times of sunrise and sun- 
set, and the right angled triangle, 
abc, or ac, will be the altitude of the 
sun, when east and west. 

The arc am corresponds to the times 
after and before six, when the sun sets and rises, and an is 
the arc on the horizon towards the north from the east and 
west points, when the sun rises and sets. 

In short, the solution will be as in the preceding exam- 




SPHERICAL TRIGONOMETRY 



131 



373] 

pies, and the triangle amn in this cut will illustrate all of 
them. 

Tan. D, 23° 24' 

Tan. L, 59° 5& 

Sin. 48° 22' 32" ; am, 



9.636226 
10.237394 



9.873620 



In time, 



3* 13 OT 30 s . 



Adding this interval to 6 hours, gives the time of sunset, 
and subtracting it from 6 hours, will give the time of sun- 
rise. 

For the point n on the horizon, we have 

Sin. man, or cos. lat. : sin. D. : : R. : sin. an. 
iZ.sin. D 19.598952 

9.699844 



Sin. an- 



cos. L 
N. of E., sin. 52° 26' 18" 



9.899108 

In the right angled spherical triangle abc, we have the 
angle EaZ, the latitude, and be, the declination. 

Sin. L. : sin. be, 23^ 24' : : B. : sin. ac. 
ac is the altitude of the sun when the sun is at the point 
c, east or west on the prime vertical. 

Jg.sjp. D 19.598952 

sin. L 9.937238 



Sin. Alt, 



Sin. ac, 27° IS' 57" 



9.661714 



To find ba, the time before and after six o'clock, apparent 
time, when the sun is east and west, we have 
R. : cos. ab : : cos. be : cos. ac. 

jg.cos. ac 19.948653 

Cos - ab ~ cos. be 9.962727 

Cos. ab, 14° 30' 30" 9.985926 

This arc (14° 30' 33") reduced to time, is equivalent to 
58 minutes 2 seconds. Hence the sun is east at 6 h 58 m 2 s , 
A.M., and it is west at 5 h l m 58% P.M., in latitude 59° 56' 



132 



KEY TO 



[377 



when the sun's declination is 23° 24' north. Problem (2,) 
[page 376]. 

By the formula on same page (Text-book), we must operate 
as follows, 

36° 12' 

40° 21' 

93° 20' 



True Alt., 

Lat. north, 

North polar dis. 



cos. com: 0.117986 
sin. com. 0.000735 



2)169° 53' 



84° 56' 30" 
36° 12' 



COS. 



48° 44' 30" sin. 



8.945320 

9.876070 
2)18.940111 
Sin. iP, 17° 10' 1" 9.470055 

The double of this angle, or (34° 20' 2"), changed into time, 
is 2 h 17 /n 20*, the interval of time from apparent noon. 

Whence, from noon, 12 7 > m s 



Take, 


2 h VJ m 


'20 s 






g h ^2 m 


40 s A.M., apparent time. 


(3.) Alt., 

Latitude, 

South polar dis. 


40° 8' 

21° 2' 

108° 32' 


cos. com. 
sin. com. 

cos. 
sin. 


0.029945 
0.013228 




2)169° 42' 
84° 51' 

40° 8' 


8.953100 




44° 43' 

sin. 15° 29' 40" 

2 


9.847327 
2)18.853500 


2-* 5 


9.426750 



30° 59' 20" =2 h 3 m 57 s , P.M. Ans. 




•$7 7] SPHERICAL TRIGONOMETRY. 133 

A Geographical Problem, (page 377, Text-book.) 
Kequired, the number of degrees on a great circle between 
New Orleans and Kome, also the number 
of miles taking 69.16 miles to each de- 
gree. 

Let P, be the north pole on the earth, 
JSf, the position of New Orleans, and B 
the position of Kome, which positions 
are 

Latitude. Longitude. 

New Orleans, 29° 57' 30" N. 90° W. 

Kome, 41° 53' 54" N. 12° 28' 40" E. 

The co. latitude of N. 0. is A 7 P=60° 2' 30". 

co. latitude of Kome is PB=48° 6' 6". 
And the sum of longitudes is 102° 28' 40", which is the 
angle NPB. The supplement of the angle NPB, is BPD, 
77° 31' 20". BD, is the perpendicular let fall on NP pro- 
duced. 

Now, in the triangle PBD, we have 

B. : sin. PB, 48° 6' 6" : : sin. 77° 31' 20" : sin. BD. 

Sin. PB 9.871767 

Sin. P 9.989619 



Sin. BD, 46° 36' 53" 9.861386 

P.cos. PB= cos. PP. cos. BD. 

„ __ P.cos. PP. 19.824654 

Cos. PD= 777.- 

cos. BD 9.836894 

Cos. PD=cos. 13° 32' 21" 9.987760 
Add NP, 60° 2' 30" 



Sum, or ND, 73° 34' 51" 



134 KEY TO l&tg 

In the right angled triangle NDR, we have 
i^.cos. NB—cos. ND. co&. DR. 
Cos. ND, 73° 34' 51"" 9.451268 

Cos. BR, 46° 36' 53" 9.836894 



Cos. NR, 78° 48' 15" 9.288162 ■ 

That is, the distance in degrees is 78°.8041, and at the rate 
of 69.16 miles to each degree, gives 5450.1 English miles. 

We can compute this side more concisely by using the 
formula 

Cos. c=cos. (a + Z>)+2cos. 2 |C.sin. a.sin. b. 

Let the angle NPR= 0=102° 28' 40", a=60° 2' 30", and 
5=48° 6' 6", c-the arc NR. 

Nat. Cos. (o + 6)=5Tat. cos. 108° 8' 36"=:— .31139. 
Log. 2 0.301030 

Cos. \Q Z R-l -1.796627 

-1.796627 
. Sin. a -1.937712 

Sin. b -1.871767 

Nat. number, 0.50555 -1.703763. Sum. 

Add, -0.31139 

Nat. cos. 78° 48' 15" .19416. Ans. same as before. 

Ncte. — There are nine examples on page 319, Text-book ; we shall show 
the solutions of only two or three of them, by a special formula, which we 
think the most concise, all things considered. Others may solve them by this 
or any one of the other methods, explained in the Geometry. 

The difference of the right ascensions of the two bodies at 
the time designated, changed into arc, will be the included 
angle of a spherical triangle, and the complements of the 
declinations are the two sides of such a triangle ; the third 
side is required and we designate it by x. The angle at the 
pole of the celestial equator, we designate by P, and a and b 
the sides. 



379] SPHERICAL TRIGONOMETRY. 135 

Now, by the Fundamental Equation in Spherical Trigo- 
nometry, we have. (See page 342, Geom.) 
_. _ cos. x — cos. a.cos. b 

COS. JP= = : j . (1) 

sin. o.sin. 6 v J 

Let D be the greater declination, and d the less. 
Then D, and d, are complements of a and b, and cos. a= 
sin. D, and sin. a=cos. D, and the above equation becomes, 

~ _ cos. x— sin. -D.sin. c£ 

Cos. P— T\ j • 

cos. JJ.cos. d 

Subtracting each member from unity, we obtain 

._. „ cos. x— sin. 2>.sin. c£ 

1— cos. P—l — ^ . . 

cos. ZJ.cos. a 

- _ (cos. D. cos. cZ-f-sin. D. sin. d)— cos. a?. 

1— cos. P~ = >- ■. 

cos. lAcos. a 

That is, 2 sin.' iP= J^S=ft=S^E , 

cos. lAcos. a 

Whence, Cos. x=cos. (D— d)~ 2sio. 2 jP.coa. £>. cos. d (F) 
This formula is essentially the same as that in the Text- 
book. 

In that, the given sides of the triangle are used. In this, 
it is the complements of those sides. 





E X A 


UPLES. 






June 24, 1880. . 


A.t noon, mean 


time. 




B. A. 




Declination. 


Moon, 


10* 5l m 36.5 s 


d= 


3° 35' 24" N. 


Jupiter, 


S h 4"' 27.6 s 
2 h 47" 8.9 s 


D = 

(D-d) = 


20° 51' 36.8" N. 


Diff. 


17° 16' 12.8" 


Or, 


41° 47' 13.5" 


=P 






20° 53' 36.8" 


= iP. 





We now apply the formula (F). 



136 key to [379 

Log. 2 0.301030 

Log. sin. \P, 20° 53' 36.8" -1.552221 

Log. sin. \P, 20° 53' 36.8" -1.552221 

Log. cos. D, 20° 51' 36.8" -1.970557 

Log. cos. d, 3° 35' 24" —1.999147 

Nat, num. " 0.237233" -1.375176 

Nat. cos. (D-d) = 0.954915 

Diff.=Nat. cos. x= 0.71768^=44° 8' 12". Ans. 

Notes. — 1. The three remaining problems in the same group are solved in 
the same manner. 

2. Let the student observe that if the natural sine or cosine is required to 
more than 5 decimal places, the logarithmic sine or cosine should first be taken 
out; and from this, diminishing the index by 10, the number may be obtained 
correctly to G or 1 places. Conversely, if the arc is required from the natural 
sine or cosine, first find the logarithm, then the arc. 

We shall now solve one or two in the next group, where the 
distances are greater than 90°, and the declinations on oppo- 
site sides of the equator. The formula employed is the same. 
Recollect, however, that the cosine of an arc, greater than 90°, 
must be taken with a minus sign. 

Example 1st., October 6, 1860. At noon. 





JR. 


A. 




Declination. 


Sun 


Yl h 49 


m 29.3* 


-- 


-d 5° 18' 42.6" S. 


Moon, 


5 h 41 


m 20.8 9 


-a) 


D 26° 8' 00 K 




T 8 


m 8.5 s (D- 


= 31° 26' 42.6" 


In arc, 


107° 
53 d 

Log. 2 


2' 7.5"=P 
31' 3.8" = ±P. 




0.301030 




Log. sin. 


53° 31' 3.8" 




-1.905278 




Log. sin. 


53° 31' 3.8" 




-1.905278 




Cos. 


26° 8' 




-1.953166 




Cos. 

,N. 


5° 18' 42.6" 




-1.998130 


Nat 


-1.155798 


-0.062882 


Nat. cos. 


(D-d) 

. COS. XAJ. 


0.853139 


107 : 




Diff. 


-0.302659 = 


d 37' 2". 



379] SPHERICAL TRIGONOMETRY. 137 

We will work the next example in this group by Equa- 
tions (8) and (9), page 350, Text-book. 

Making G the included angle, and A, the angle opposite 
the greater polar distance. 

®RA, 12* 49™ 56.7 s Dec. 5° 21' 35.4" &. 
®BA, 5 h 48™ 30.1 s Dec. 26° 3' 20" N. 

In degrees, 105° 21' 39" = Q 

52° 40' 49.5" =i<7. 

The distance from the north pole to the center of the sun, 
is the sun's declination added to 90°, which we designate by a. 
Therefore, a =95° 21' 35.4" 

The moon from the same point is b — 63° 56' 40" 
Half sum, is 
Half diff., is 
Cot. \C, 



X 

x 

9.882147 

9.983467 


(a + b) =79° 39' 

(a-b) = 15° 42' 

cot. i(7, 

sin. l(a— b) 

sin. ±(a + b) 


7.7" 
27.7" 

9.882147 
9.432536 


19.865614 
9.254364 


19.314683 
9.992878 



Cos. \{a—b) 

Cos. \(a + b) 

Tan. i(A+B), 10.611250 tan. l(A—B) 9.321805 
%(A+B), 76° 14' 47" HA-B) 11° 50' 56 n 
l(A-B), 11° 50' 5& 1 

Diff.=5=64° 23' 51" 

Lastly, for the side sought, we have the proportion 
Sin. 5, 64°23'51" : sin. 5, 63°56'40" : : sin. 105°21'39" : sin. x. 
Sin. b 9.953454 

Sin. C= cos. 15° 21' 39" 9.984201 

19.937655 
Sin. B 9.955116 

Sin. 106° 8' 19"=cos. 16° 8' 19" 9.982539 

We prefer the other formula. 



138 KEY TO 



ASTRONOMICAL PROBLEMS. 

The following Astronomical problems were included in 
Robinson's Geometry, as first published, but being deemed 
too difficult for such a work, were omitted in the New 
Geometry. 

(1.) In latitude 40° 48' north, the sun bore south 78° 16' 
west, at 3 h 38™ P.M. ; apparent time. Required his altitude 
and declination, making no allowance for refraction. 

Ans. The altitude, 36° 46', and declination, 15° 32' north. 

z Let Hh be the horizon, Z the 

zenith of the observer, P the north 
pole, and PS a meridian through 
the sun. 

PZ is the co-latitude, 49° 12', and 
PS is the co-declination or polar 
distance, one of the arcs sought. 
ZS is the co-altitude, or ST is the altitude of the sun at 
the time of observation. 

The angle ZPS is found by reducing 3 ;i 38™ to degrees 
at the rate of 4 m to one degree ; hence, ZPS =54° 30'. 

Because HZS=78° 16', PZS '=101° 44'. From Z let fall 
the perpendicular ZQ on PS. Then in the right angled 
spherical A PZQ, equation (13) gives us* 

iZ.sin. ZQ=sin. PZ sin. P 

sin. P^=sin. 49° 12' 9.879093 

sin. P = sin. 54° 30' 9.910686 

sin. ZQ=sm. 38° 2' 42" 9.789779 

* To apply the equations without confusion, letter each right angled spheri- 
cal triangle ABC, right angled at B, then A must be written in place of P ; and 
when operating on ZSQ, write A in place of S, and G for the angle SZQ. 




ASTRONOMICAL PROBLEMS. 139 

To obtain the angle PZQ, we apply equation (19), which 
gives 

JB.COS. PZQ=cot PZ.Uu. ZQ. 
That is, P.cos. P^Q=tan. 40° 48'.tan. 38° 2' 42". 
Tan. 40° 48' 9.936100 

Tan. 38° 2' 42" 9.893513 



PZQ=cos. 47° 30' 30" 9.829613 

PZS= 101° 44' 



SZQ= 54° 13' 30" 
To obtain ZS or its complement, we again apply (19) 

(19) P.cos. SZQ=cot. ZSten. ZQ. 
That is, P.cos. 54° 13' 30" = tan. STten. 38° 2' 42". 
P.cos. 54° 13' 30"= 19.766761 

tan. 38° 2' 42"= 9.893513 



Tan. 36° 46', nearly 9.873248 

To find PS, we take the following proportion, 

Sin. P : sin. ZS : : sin. PZS : sin. PS. 
That is, 
Sin. 54° 30' : cos. 36° 46' : : sin. 101° 44' : sin. PS. 
Cos. 11° 44' 9.990829 

Cos. 36° 46' 9.903676 



19.894505 
Sin. 54° 30' 9.910686 



PS, sin. 74° 28' 9.983819 

Whence, the sun's distance from the equator must have 
been 15° 32' north. 

(2.) In north latitude, when the sun's declination was 
14° 20' north, his altitudes, at two different times on the 



140 KEY TO 

same forenoon, were 43° 7' + ? and 67° 10'+ : and the change 
of azimuth ; in the interval, 45° 2'. Bequired the latitude. 

Am. 34° 21' 14" north. 

....-£. Let PK be the earth's axis, Qq the 

/\yy— ^<n p equator, and IZ/i the horizon. 

/ / y§ /^ \ Also, let Z be the zenith of the ob- 

\ **~^^r — —j:/i server, Son the first altitude, Tw- the 

'^/ \V second, and the angle TZS=45° 2'. 

\.. ,..--''2 Our first operation must be on the 

triangle zm ZT=22° 50', ^#=46° 53', and we must find 

TS, and the angle TSZ. 

From I 7 , conceive TB let fall on ZS, making two right 
angled a's ; and to avoid confusion in the figure, we will 
keep the arc TB in mind, and not actually draw it. 
Then the A ZTB furnishes this proportion, 
B : sin. 22° 50' : : sin. 45° 2' : sin. TB=sm. 15° 5& 8". 

To find ZB we have the following proportion, 

B : cos. ZB : : cos. 15° 5® 8" : cos. 22° 50'. 

Whence, we find ^5 = 16° 34' 13". Now in the right 
angled spherical A TBS,. we have TB=15° 5& 8", BS= 
4G°53'-16°34'13", or BS=30° 18' 47" ; and TS is found 
from the following proportion, 

B : cos. 15° 56' 8" : : cos. 30° 18' 47" : cos. TS. 

This gives TS=33° 53' 26". To find the angle TSZ, we 
have the proportion, 

Sin. 33° 53' 26" : B : : sin. TB 15° 56' 8" : sin. TSZ. 

Whence, the angle TSZ=29 D 29' 49". 

The next step is to operate on the isosceles spherical A 
PTS. We require the angle TSP. 

Conceive a meridian drawn bisecting the angle at P, it will 



PROBLEMS. 141 

also bisect the base TS, forming two equal right angled spher- 
ical triangles. 

Observe that PS =7 '5° 40' and J TS=16° 5& 43". 

To find the angle TSP, we apply equation (19), in which 
a =16° 5& 43", 5 = 75° 40', and the equation becomes, 

i^.cos. TSP-cot. 75° 40'.tan. 16° 56' 43". 
Whence, TSP=85° 32' 5", and PSZ=85° 32' 5"-29° 29' 49" 
=56° 2' 16". 

The third step is to operate on the A ZSP ; we now 
have its two sides ZS and SP, and the included angle. 

From Z conceive a perpendicular arc let fall on SP, call- 
ing it ZB ; then the right angled spherical triangle SZB, 
gives 

R. : sin. ZS : : sin. ZSB : sin. ZB. 

That is, 
B. : sin. 46° 53' : : sin. 56° 2' 16" : sin. ZB = sin. 37° 15' 37". 

To find SB, we have the following proportion, 
B. : cos. SB : : cos. ZB : cos. ZS. 

That is, B. : cos. &£ : : cos. 37 D 15' 37" : cos. 46° 53'. 

Whence, #£=30° 49' 18". Now, from PS, 75° 40', take 
SB, 30° 49' 18", and the difference must be JSP, 44° 50' 42". 

Lastly, to obtain PZ, and consequently ZQ the latitude, 
we have 

B. : cos. ZB : : cos. BP : cos. ZP=sm. ZQ. 

That is, B. : cos. 37° 15' 37" : : cos. 44° 50' 42" : sin. ZQ= 
sin. 34° 21' 14" north. 

This is the result by a careful computation, and it differs 
1' 14" from the answer given in the text-book. 

This is a modification of latitude by double altitudes, but 
in real double altitudes the arc TS is measured from the 
elapsed time between the observations, and the angle TZS 
is not given. 



142 



ASTRONOMICAL 



(3.) In latitude 16° 4' north, when the sun's declination is 
23° 2' north. Kequired the time in the afternoon, and the 
sun's altitude and bearing when his azimuth neither increases 
nor decreases. 

Ans. Time, 3* 9 m 26 s P.M., altitude, 45° 1', and bearing 
north 73° 16' west. 

Let Pp be the earth's axis, Hh 
the horizon, Qq the equator, QZand 
Ph, each equal to 16° 4' north, and 
Qd, qd, each equal to 23° 2' ; then 
the dotted curve dd represents the 
parallel of the sun's declination. 
Through Z and N an infinite 
number of vertical circles can be drawn, one of these will 
touch the curve dd ; let it be ZON. 

At the point where this circle touches the curve dd will 
be the position of the sun at the time required, and POZ 
will be a right angled spherical A , right angled at 0. The 
problem requires the complement of ZO, and the time cor- 
responding to the angle ZPO. 

In the spherical A POZ, we have 

Pi. : cos. PO : : cos. ZO : cos, PZ 
That is, P. : sin. 23° 2' : : sin. altitude : sin. 16° 4'. 
P.sin. 16° 4' 




Whence, sin. alt. 



, = sin. 45° 1' nearly. Ans. 

sin. 23° 2' J 



To find the angle at P, we have the following proportion, 
Cos. 16° 4' : P. : : cos. 45° 1' : sin. P. 

Whence, sin. P=sin. 47° 21' 40", and ZPO=47° 21' 40", 
which being changed into time, at the rate of 15° to one 
hour, gives 3 h 9 m 26 s . 

To find the angle PZO, we have the proportion, 

Cos. 16° 4' : B. : : cos. 23° 2' : sin. P^O=sin. 73° 16 r . 



PROBLEMS 



143 




(4.) The sun set south-west J south, when his declination 
was 16° 4' south. Kequired the latitude. Ans. 69° 1' north. 

Draw a circle as before. Let Hh z 

be the horizon, Z the zenith, P the 
pole. The great circle PZH is the 
meridian, and ZCN at right angles to 
it, and of course east and west. Let 
BC be a portion of the equator, and 
BO the arc of declination. The posi- 
tion on the horizon where the sun set is the arc HO=45°— 
5°37'30":=39 o 22'30". 

Consequently, the arc OC=50° 37' 30". 

In the right angled spherical triangle BOO, we have BO, 
BO given to find the angle BOO, which is the complement 
of the latitude, or the complement of the angle BCZ. 

To find the angle BCO, we apply equation (14). 
iZ.sin. BO=sin. 00 An. BOO. 

That is, i?.sin. 16° 4' = sin. 50° 37' 30".sin. BOO. 
E.sin. 16° 4' 19.442096 

Sin. 50° 37' 30" 9.888186 

Cos. 69° 1' nearly 9.553910 

Scholium. — The arc BO on the equator measures the 
angle BPO, corresponding to the time from 6 o'clock to sun- 
rise or sunset. This arc is called the arc of ascensional dif- 
ference in astronomy. The time of sunset is before six, if 
the latitude is north and the declination south, as in this 
example, but after six, if the latitude and declination are 
both north or both south. 

To obtain this arc, the latitude and declination must be 
given ; that is, BO and the angle BCO, the complement of 
the latitude. Here we apply (12), that is, 
E.sin. BC= tan. D.tan. Z, 



144 ASTRONOMICAL 

an equation in which D represents the declination, and L the 
latitude. 



(5.) The altitude of the sun ; when on the equator, was 
14° 28'+ bearing east 22° 30' south. Kequired the latitude 
and time. Ans. Latitude 56° 1', and time 7 h 46 m 11*, A.M. 

Let S he the position of the sun on the equator. (See the 
last figure.) Draw the arc ZS, and the right angled spheri- 
cal A ZQS is the one we have to operate upon. 

Then ZS is the complement of the given altitude, and the 
angle QZS, is the complement of 22° 30'. The portion of 
the equator between Q and S, changed into time, will be the 
required time from noon, and the arc QZ will be the required 
latitude. 

First for the arc QS. 

R : sin. ZS : : sin. QZS : sin. QS. 

That is, 

B. : cos. 14° 28' :: cos. 22° 30' : sin. QS=63° 27' 19". 

But 63° 27' 19" at the rate of 4 m to one degree, corres- 
ponds to 4 h 13™ 49 s from noon — and as the altitude was 
marked + , rising, it was before noon, or at 7 h 46 m 11 s in the 
morning. 

To find the arc QZ, we have the following proportion, 
R. : cos. 63° 27' 19" : : cos. QZ : sin. 14° 28'. 

Whence, cos. QZ— cos. 56° V nearly, and 56° V is the lat- 
itude sought. 

(6.) The altitude of the sun was 20° 41' at 2 h 20 m P.M. 
when his declination was 10° 28' south. Kequired his 
azimuth and the latitude. 

Ans. Azimuth south 37° 5' west, latitude 51° 58' north. 



PROBLEMS 



145 




This problem furnishes the spheri- 
cal aPZO, in which the side ZO is 
the complement of 20° 41' or 69° 19'. 
P0=90 o 4-10 o 28' = 100° 28', and the 
angle ZPO is 2 h 20™ changed into 
degrees at the rate of 15° to one 
hour, or ZPO = 35°. 

Now in the triangle ZPO, we have 

Sin. ZO : sin. ZPO : : sin. PO : sin. PZO. 
That is, 
Cos. 20° 41' : sin. 35° : : cos. 10° 28' : sin. P^O=sin. 37° 5'. 

In the right angled spherical A BOZ, Ave apply equa- 
tion (16). 

(16). P.cos. 37° 5'= tan. 20° 41'.tan. BZ. 
iZ.cos. 37° 5' 19.901872 

Tan. 20° 41' 9.576958 

Tan. £Z=tan. 64° 40' 10.324914 

To find PB in the right angled A BPO, we apply the 
same equation (16). 

P.cos. 35°=tan. 10° 28' tan. PB. 
P.cos. 35° 19.913365 

Tan. 10° 28" 9.266555 

Cot. 12° 42' 10.646810 

But PB is obviously greater than 90°, therefore the point 
B is 12° 42' below the equator ; but from B to Z is 
64° 40' ; therefore, from Z to the equator, or the latitude, 
is the difference between 64° 40' and 12° 42', or 51° 58' 
north. Ans. Lat. 51° 58' north. 



(7.) If in August 1840, Spica was observed to set 2 h 26 m 
14 s before Arcturus, what was the latitude of the observer, 



146 ASTRONOMICAL 

no account being taken of the height of the eye above the sea, 
nor of the effect of refraction ? Ana. 36° 47' 38" north. 

By a catalogue of the stars to be found in the author's 
Astronomy, or in any copy of the English Nautical Almanac, 
we find the positions of these stars in 1840, to have been as 
follows : 

Spica, right ascension, 13 /J 16 m 46 s Dec. 10° 19' 40" south. 
Arcturus, " " 14* 8 m 25 s Dec. 20° 1' 4" north. 

Let L— the latitude sought. Put d=10° 19' 40", and 
Z>=20°1'4". 

The difference in right ascensions is 51 m 39 s , and this 
would be about the time that Arcturus would set after 
Spica, provided the observer was near the equator or a little 
south of it ; but as the interval observed was 2 h 26 m 14 9 , the 
observer must have been a considerable distance in north 
latitude. In high southern latitudes Arcturus sets before 
Spica. 

When an observer is north of the equator, and the sun or 
star south of it, the sun or star will set within six hours 
after it comes to the meridian. 

When the observer and the object are both north of the 
equator, the interval from the meridian to the horizon is 
greater than six hours. 

The difference between this interval and six hours, is called 
the ascensional difference, and it is measured in arc by BO 
in the figure to the 4th example. 

Now let x— the ascensional difference of Spica corres- 
ponding to the latitude L, and y= that of Arcturus corres- 
ponding to the same latitude ; then by the scholium to the 
4th example, calling radius unity, we shall have 

sin. cc=tan. Z.tan. d (1) 

sin. y= tan. Z.tan. D (2) 



PROBLEMS. 147 

The star Spica came to the observer's meridian at a cer- 
tain time, that we may denote by M. 

Then 31+ ( 6 — — ) = the time Spica set. 

And M+ 51 m 39 s + (6 + |Q = the time Arcturus set. 

By subtracting the time Spica set from the time Arcturus 
set, we shall obtain an expression equal to 2 h 26 m 14 s . 

That is, 51 m 39 s +^ + ^=2" 26™ 14 s . 

Or, g+g=P34-3^ (3) 

x+y=15 (P 34™ 35 s ) ( 4 ) 

Equation (3) expresses time. Equation (4) expresses arc. 
When we divide arc by 15, we obtain time, one degree 
being the unit for arc, and one hour the unit for time ; there- 
fore, when we multiply time by 15, we obtain arc ; that is, 
1* multiplied by 15, gives 15° ; hence (4) becomes 
x + y=23°38 , 45" = a 
x=a—y (5) 

That is, the arc x is equal to the difference of the arcs a 
and y ; but to make use of these arcs and avail ourselves of 
equations (1) and (2), we must take the sines of the arcs, 
(see equation (8), plane trigonometry) ; then (5) becomes 
Sin. x=sin. a. cos. y—cos. a. sin. y. (6) 

Substituting the values of sin. x and sin. y from (1) and 
(2), (6) becomes 
Tan. L. tan. d— sin. a. cos. y—cos. a. tan. L. tan. D (7) 
Squaring (2), sin. 2 ?/ = tan. 2 Z. tan. 2 D. 

Subtracting each member from unity, and observing that 
(1— sin. 2 ?/) equals cos. 2 ?/, then 

Cos. 2 ?y=:l-tan. 2 Z. tan. 2 Z>. 



Or, Cos. y = i'l-tan.'.L. tan. 2 Z>. 



148 ASTRONOMICAL 

This value of cos. y put in (7), gives 



Tan..L.tan.d=sin.a. t 1— tan. 2 .L.tan. 2 Z>— cos.a.tan.Z.tan.Z) (8) 
By transposition and division, 

/T an .rf + cos.a.tan.m. tan , L = Vl-tan.'I, tan.<i> 
\ sin. a / 

Squaring, 



/Ian. d-\- cos. a. tan. ZA" 27 -_ 
\ sin. a / ' ' ~~ 



l-tan.' J Z.tan. 2 X> 



Dividing by tan. 2 Z, and observing that -=cot. 2 Zr we 

, /Tan. d+cos. a. tan. Z>\ 2 r 2 

have ( : I =cot."Z— tan. i? 

\ sin. a / 

Or, Cot,'Z = tan.W ta n - d+cos - a - tan ' D \* 

' sin. a / 

, 271 , /tan. cZ tan. D\ 2 

= tan. 2 Z> + _}- I 

\ sm. a tan. a' 

We must now find the numerical value of the second 
member. Using logarithmic sines, cosines, tangents, &c, 
we must diminish the indices by 10, because the equation 
refers to radius unity. 
Log. tan.2?. = -1.561485; tan. 2 _D= -1.122970=0.132730 num 

Log. tan. d -1.260623 log. tan. D -1.561485 

sin. a -1.603233 tan. a. -1.641318 



0.454349 -1.657390 0.832083 -1.920167 

0.454349 + 0.832083=1.286432 (1.286432) 2 = 1.654906 
Whence, cot. 2 £ = 0.132730 + 1.654906=1.787636 

Square root, cot. L= 1.337025 

Taking the log. of this number, increasing its index by 10 
will give the log. cot. in our tables. 
Log. 1.337025=0.126139 + 10. = 10.126139=cot. 36° 47' 38". 

(8.) On the 14th of November, 1829, ' Menkar was ob- 



PROBLEMS. 149 

served to rise 48™ 3 s before Aldebaran : what was the lati- 
tude of the observer. Am. 39° 33' 53" north. 

The position of these two stars in the heavens, November 
1829, were as follows : 

Menkar, right ascension, 2 h 53™ 21 s . Dec. 3° 24' 52" north. 
Aldebaran, " 4 h 26™ 7 s . Dec. 16° 19' 31" north. 

Aldebaran passes the meridian l k 32™ 46 s after Menkar. 
Now let M represent the time Menkar was on the meridian, 
then M + l h 32™ 46 s represents the time Aldebaran was on 
the meridian. Also, let x= the arc of ascensional difference 
corresponding to the latitude and the star Menkar, and y 
that of the star Aldebaran. 

Then, M—fo +~)= the time Menkar rose. 

And, if + l h 32™ 46 s — ( 6 + JL \ = the time Aldebaran rose. 
Subtracting the upper from the lower, the difference must 

:48™ 3 s 

= -0.74527. 

That is, l h being the unit, 44™ 43*= 0.74527 of an hour, 
and multiplying by 15, we shall have as many degrees of arc 
as we have units ; therefore, 

x-y= -(0.74527)15= -11° 10' 45"= -a 
x—y—a 
sin. cc=sin. i/.cos. a— cos. y.sin. a (1) 

Put d=3° 24' 52", D=16°19'31", and Z = the required 
latitude. Then by scholium to the 4th example, 

Sin.o: = tan. d. tan.X. sin.?/ = tan.Z). tan.iv. 



be 48™ 3 s ; that is, 




1* + 32™ 46 s - 


y x 

~15 + 15 


Whence, -— — $- = 
15 15 


—44™ 43 



150 ASTRONOMICAL 

The values of sin. x and sin. y, substituted in (1), give 

Tan. dtan. L=cos. a.tan. D.tan. L— cos. y.sm. a (2) 

But, sin. 2 2/=tan. 2 J9.tan. 2 .L, and 1— sin. 2 ?/=l— tan. 2 D.tan. 2 £. 

Or, cos. 2 ?/=l — tan. 2 Z>.tan. 2 .L. 

Or, cos. y=Vl— tan 2 i>.tan. 2 iv. 

By substituting this value of cos. y in (2) and transposing, 
we find 



Sin.a^l— tan. a l?.tan. a i/= (cos. a.tan. B. — tan. d)tan. L: 

cos ct 

Dividing; bv sin. a, and observing that . ' = , we have 

& J ' ° sin. a tan. a' 

./i — i — on. 2 T /tan. 2) tan.cA, r 

yl— tan.-Z>.tan. 2 £ = (- — )tau.L. 

Vtan.a sin. a J 

Squaring and dividing by tan. 2 £, and at the same time ob- 
serving that ==cot. L. and we shall have 

° tan. L 

n xor j. *r> /tan.Z) tan.aV 

Cot. 2 £— tan. 2 Z> = ( : J 

nan. a sm.a/ 

We will now find the numerical value of the known quan- 
tities. 

Log. tan. D -1.466718 log. tan. cl -2.775712 

Log. tan. a -1.295849 log. sin. a -1.287530 

Log. 1.482072 0.170869 log. 0.307738 -L488182 

Tan. 2 Z> = 0.085790 1.482072 - 0.307738 = 1. 174334 
Whence, cot. 2 i: -0.085790 = (1.174334)'. 

Or, cot. 2 i: = 1.464849. 

cot. £=1.210309. 
Log. cot. £ + 10=10.082896 = cot. 39° 33' 53". Ans. 



(9.) In latitude 16° 40' north, when the sun's declination 
was 23° 18' north, I observed him twice, in the same fore- 



PROBLEMS. 



151 



noon, bearing north 68° 30' east. Kequired the time of ob- 
servation, and his altitude at each time. 
Ans. Times, 6 h 15 m 40 s , A.M., and 10* 32™ 48 s , A.M., alti- 
tude, 9° 59' 33", and 68° 29' 43". 

Let Z be the zenith, P the north 
pole, and the curve dd be the parallel 
of the sun's declination along which 
it appears to revolve. Make the an- 
gle PZS' equal to 68° 30' ; then the 
sun was at S at the time of the first 
observation, and at S' at the time of 
the second. 

In the spherical A PZS' there is given PZ, PS' and the 
angle PZS' ; also, in the A PZS there is given PZ, PS, 
and the angle PZS. Observe that PSS' is an isosceles A . 

Describe the meridian PB bisecting the angle SPS, and 
then we have three right angled spherical triangles, BPS, 
BPS, and BPZ ; taking the last, we have the following 
proportion : 

P. : sin. PZ : : sin. PZB : sin. PB. 
That is, 
B. : cos. 16° 40' : : sin. 68° 30' : sin. PB=sin. 63° 2' 27". 




To find ZB, we take the following proportion, (see page 
185, Observation 1, Eobinson's Geometry), 

B. : cos. ZB : : cos. BP : cos. PZ. 
That is, B. : cos. ZB : : cos. 63° 2' 27" : sin. 16° 40'. 
B.sin. 16° 40' 19.457584 

Cos. 63° 2' 27" 9.656439 

Cos. ZB, 50° 45' 22" 9.801145 



152 ASTRONOMICAL 

To find S'B, we have 

B. : cos. ff'JB : : cos. 63° 2' 27" : sin. 23° 18'. 
B.sin. 23° 18' 19.597196 

cos. 63° 2'' 27" 9.656439 



Cos. S'B, 29° 15' 5" 9.940757 

Observe that S'B^BS ; therefore, ^#=50° 45' 22" +29° 
15' 5"=80° 0' 27", and ^#'=50° 45' 22"-29° 15' 5"=21 Q 
30' 17", the complements of the altitudes. Consequently the 
altitude at the first observation was 9° 59' 33", and at the 
second, 68° 29' 43". 

To find the time from noon at the first observation, we 
have the following proportion, 

Sin. PS : sm.PZS : : sin. ZS : sin. ZPS. 
That is, 
Cos.23°18' : sin.68°30' : : sin.80°0'26" :sin.^P>S'=sm.86 5 , 7" 
Had the angle been 90°, the time would have been just 
6 h , but the angle 3° 54' 53" less ; this corresponds to 15 w 40 s , 
in time. Therefore, the time was 6 h 15 m 40 9 . For the time 
at the second observation, we have 

Cos. 23° 18' : sin. 68° 30' : : sin. 21° 30' 17" : sin. ZPS' 

[=sin.21°47'57" 
21° 47' 57" =1* 27™ 12 s from noon, or 10* 32 w 48 s apparent 
time in the morning. 



(10.) An observer in north latitude marked the time when 
the stars Kegulus and Spica were eclipsed by a plumb line, — 
that is, when they were both in the same vertical plane passing 
through the zenith of the observer. One hour and ten min- 
utes afterwards, Kegulus was on the observer's meridian. 
What was the observer's latitude ? 




PROBLEMS. 153 

The positions of the stars in the heavens were 

Regulus, right ascension, 1Q* m 10*. Dec. 12° 43' north. 

13*17™ 2 s . Dec. 10° 21' 20" south. 

Let B be the position of Regulus, S 
the position of Spica, P the pole, and 
Z the zenith. 

Then the side P#=100° 21' 20' f ? 
PB=77° 17', and the angle BPS= 
3 h 16 m 52 s , converted into degrees ; that 
is, BPS= 49° 13'. 

One hour and ten minutes reduced to arc, give 17° 30' ; 
but the stars revolve according to siderial, not solar time, 
and to reduce solar to siderial arc, we must increase it by 
about its aljth part ; this gives about 3' to add to 17° 30', 
making 17° 33' for the angle ZPB. Our ultimate object is 
to find PZ, the complement of the latitude. 

In the A PBS, we have the two sides PB, PS, and the 
included angle P, from which we must find BS and the an- 
gle SBP, and we can let a perpendicular fall from B on to 
the side PS and solve it by the usual way ; but to show 
that a wide field is open for a bold operator, we will put 
the unknown arc BS—x, the side opposite B=r, and oppo- 
site S=s, and apply one of the equations in formula (S), 
page 191, Robinson's Geometry. 

mi . , . -r, cos.sc— cos.r. cos. s 

That is, cos. P— = -= . 

sin. r. sin. s 

Whence, Cos. P.sin. r.sin. s + cos. r.cos. s=cos. x. 

We now apply this equation, recollecting that radius is 
unity, which will require us to diminish indices of the loga- 
rithms by 10. 



-1.815046 






-1.992868 


■—cos. 


-1.254683* 


-1.989214 


COS. 


-1.342679 



154 ASTRONOMICAL PROBLEMS. 

Cos.P=cos.49°13' 

Sin. r= sin. 100° 21' 20" 
Sin. s = sin.77°17' 

0.62680 -1.797128 .03957 -2.597362 

Cos. a=0.6268— 0.03957=.58723. 
Whence, by the table of natural cosines, we find 

a;=54°2 / 20". 
To find the angle SRP or ZRP, we have 
Sin. 54° 2' 20" : sin. 49° 13' : : sin. 100° 21' 20" : sin. ZRP. 
Whence, ZRP=66 D 57' 37". 

Let fall the perpendicular RB on PZ produced, then the 
right angled spherical A PBR gives this proportion, 
R. : sin. 77° 17' : : sin. 17° 33' : sin. BB^sin. 17° 6' 22". 
To find PB y we have 

R. : cos. PB : : cos. 17° 6' 22" : cos. 77° 17'. 
Whence, PB=76° 41'. 

Now, to find the angle BBP, we have 
Sin. 77° 17' : R. : : sin. 76° 41' : sin. BRP= sin. 86° 1'. 
From PRB take PRZ, and ZRB will remain ; that is, 
From 86° 1' take 66° 57' 37", and ZBB =19° 3' 23". 
By the application of equation (12), we find that 
iZ.sin. 17° 6' 22" = tan. BZ. cot. 19° 3' 23". 
Whence, BZ=5° 48' 

And, PZ=76° 41' -5° 48' =70° 53'. 

The complement of 70° 53' is 19° 7', the latitude sought. 
By this example we perceive that by the means of a meri- 
dian line, a good watch, and a plumb line, any person having 
a knowledge of spherical trigonometry, and having a cata- 
logue of the stars at hand, can determine his latitude by 
observation. 

* Observe that r is greater than 90°, its cosine is therefore, negative in value, 
rendering the product, cos. r. cos. s, or .03957, negative. 



KEY TO 

R O B I N S O N'S 
ANALYTICAL GEOMETRY. 



CHAPTER I. 

STKAIGHT LINES. 

(Page 105.) 
Ex. 3. y=2x + 5. 

Draw the rectangular coordinate axes, XX, YY' . 



T 3 



Then, when y — 0, the 
equation, give x = — f 
= — 2\, and when x=0, 
y=^5; hence the line repre- 
sented hy the equation cuts 
the axis of Xat the distance 
— 2}, and the axis of Yat 
the distance + 5 from the 
origin or zero point. 

Assuming any conve- 
nient unit, lay it off on the 
axis several times, in both 
the positive and negative 
directions, and draw a line cutting the axis of X at —2^-, and 
the axis of Y at +5 from the origin. This will be (3, 3), the 
line required, since two points determine the position of a line. 

Ex. 4. y—— 3#— 3. 

Making y—0 -in this equation gives x——\, and x=0 
gives y——3 ; hence, this line cuts the axis of X at —1 and 




156 KEY TO [100 

the axis of Y at —3. Laying off these distances on their 
respective axes, and drawing through the points thus deter- 
mined the line (4, 4) it is that required. 

Ex. 5. 2y=3x+5, or y==fa;-f-f. 

In this equation y=0 makes ic= — f= — If, and x=0 
makes y=^=2\. The line (5, 5), drawn through the points 
in the axis ; given by these values of x and y, is that repre- 
sented "by the equation. 

Ex. 6. y=4x— 3. 

The values of x and y given by this equation, by making 
first, y=0, and then x=6, are x=%, y——3. The line is (6, 6). 

The lines of which equations 4 and 6 are the equations 
both intersect the axis of Y at the distance —3 from the 



Ex. 7. y=-2x + 3. 



Y13 



Making in this equa- 
tion y=0 and x=0 } 
successively, we find 
sc=f=li, y—3, and 
the required line is 
(7, 7). 



Ex. 8. y=2x-3. 
Proceeding with this 
as with the preceding 
equation, we getx=l± } 
y=—3; and (8, 8) is 
the line responding to 
this equation. 
In the triangle CAB since OA and OB are each equal 3, 




105] ANALYTICAL GEOMETRY. 157 

AB, the base, is equal to 6, and because 00 is perpendicular to 
the base and bisects it the triangle is isoceles. 00 is equal 
to l|by the construction. 

Ex. 9. 3x + 5y— 15=0, or y= — faj-f-3. 

The suppositions ?/=0 and #=0, made successively, give 
#=5, 2/ =3, and the line (9, 9) drawn through the point +5 
on the axis of X, and the point + 3 on the axis of Y responds 
to the equation. 

Ex. 10. 2x— 6y + 7=0, or y—\x+l. 

From this we find the intersections of the line with the 
axes of X and Y respectively to be at the distances — § and 
+ J from the origin, and (10, 10) is the required line. 

Ex. 11. x+y + 2=0, or y=—x—2. 

This line cuts both the axes at the distance —2 from the 
origin and makes with each an angle of 45°. It is the line 
(11, 11). 

Ex. 12. — x + y + 3=0, or y=x— 3. 

This line cuts both axes at the distance of 3 units from 
the origin ; the axis of X at +3, the axis of Y at —3, it 
therefore makes with each an angle of 45°. It is the line 
(12, 12). 

Ex. 13. 2x— ?/ + 4=0, or y=2x + 4. 

In 'this ?/=0 gives #=— 2, and x=0 gives y—4\ hence, 
the line (13, 13) which cuts the axis of X at —2, and the 
axis of Y at +4 is that which responds to the equation. 

If we solve any equation of the first degree between two 
variables with reference to one of the variables it will take 
either the form y=ax + b, or x=a'y + b'. Now a denotes the 
tangent of the angle that the line makes with the axis of X, 



158 



KEY TO 



[103 



and a' denotes the tangent of the angle the same line makes 
with the axis of Y, while b is the distance from the origin to 
the point in which the line cuts the axis of Y, and b ] the 
distance from the origin to the point in which the line cuts 
the axis of X. 

To construct this line we may draw through the origin a 
line making with the axis of X an angle having a for its 
tangent, and then the line drawn parallel to this through the 
point on the axis of Y at the distance b from the origin will 
be that represented by the equation. 

To make this construction 
assume any convenient unit of 
measure, and lay it off on the 
axis of X from the origin to the 
right, that is, in the positive 
direction. Through the ex- 
tremity of this unit draw a 
parallel to the axis of Y, and 
mark off on this parallel from 
the axis of X the distance a 
units, above the axis of X if a is positive, below if it is 
negative. The line which connects the origin with the point 
thus determined on the parallel will make with the axis of 
X an angle of which a is the tangent, and it is, therefore, 
parallel to the required line. If, then, we lay off the distance 
b units on the axis of F, above, or below the origin, according 
as b is positive or negative, and through the extremity of 
this line we draw a line parallel to that passing through the 
origin we shall have the required line. 

Thus, to construct the line represented by the equation 
2x—y + 4=0 (example 13), we solve it with reference to y 
and get y= 2x + 4. 

We lay off the distance OT— unity on the axis of X from 




108] 



ANALYTICAL GEOMETRY. 



159 



the origin to the right and on the 
parallel to the axis of Y, at this 
distance, mark off TT ] = two 
units above the axis of X, because 
in this example a— +2. 

Then make 0B=4, draw OV 
and through B draw BA parallel 
to OT', and the line given by the - 
equation is constructed. 

With this explanation the 
student will find no difficulty in 
applying this method of construc- 
tion to all of the above examples. 

(Page 108.) 
To prove that equations (6) and (7) are different forms of 
the same equation. 

By clearing equation (6) of fractions it becomes 

yx u — y'x" — yx' + y'x' = y"x— y'x— y"x } -f ijx\ 
Canceling and transposing we find 




yx"—yx ] =y'x"-\-y ] 'x—y'x—y ] 



(8) 



Equation (7) treated in the same way gives, first, 

yx"—y"x !1 —yx' + y"x'=y' } x—y'x—y"x" + y'x". 
And finally 

yx 11 —yx 1 =y*x n + y"x—y'x—y"x 1 . (9) 

Since equations (8) and (9) are the same equations, (6) and 
(7) from which they were derived must represent the same 
line. 

Equation (6) being 

J u rf'—x' K J 

the suppositions y=i/, and x—x' reduces both members to 
zero; hence the line passes through the first point of which 
the co-ordinates are y 1 and x\ 



160 key to [109, 119 

If in this equation we suppose y=y" and x=x ,! it becomes 



u u x"—x ] v } 

Dividing both members of this last equation by y u —y' and 
clearing of fractions it reduces to 

x^—x ] =x r] —x ] or 0=0. 
Therefore, the co-ordinates of both points when substi- 
tuted for the variables x and y in equation (6) satisfy that 
equation, which is the condition that the line shall pass 
through these points. 

(Page 109.) 
Ex. 2. Placing in the equation 

y~y ] =y~^^-^ ( a ) 



for x\ x", and y', y" their values, this equation becomes 

JUL 4-1 

Eeducing 

2/+l = -/T*-if. 
Or, y = -j\x-lH. 

Ex. 3. Making the substitutions in equation ( a ), (last 

example) for x\ x" 7 and y 1 , ?/", it becomes 

Or, jr— 5 = fo?— V- 

Whence, «/ "= f x + 3|. 

(Page 112). 
Ex. 2. The equations 

2y = 5x + 8 
3^ = — 2x -f 6, 



115] ANALYTICAL GEOMETRY. 161 

become by dividing through by the coefficients of y, 
y = | x + 4 
y=-\x + 2. 

| the formula. 



Making 


a = | and a' = - 




a'— a 

m = r 

1 + aa' 


Becomes 






m = =±=l = 4i 



4.75. 



(Page 114.) 
Ex 1. By making a=— 2, a'=5, b=l and &'=10, in 



the formula 



x — it becomes 

a— a' 

1-10 

The same substitutions in 

db—aV . 

5 + 20 _ 2 

y '~ 5+2"-" T "- dT * 



(Page 115.) 

Ex. 1. In the formula for the perpendicular, 

-p. , b + ax'—y' 
Per. = ± ^ JL , 

we must make a = 3, & = — 10, as' = 4, and 2/'=5. These 
values being substituted, we have 

Per . =± =10+12=5 -3_ 

V3* + l 4/10 

Multiplying the numerator and denominator of this by 



162 KEY TO [H8 

VlO and passing the 3 under the radical by squaring, we 
have finally 

Perpendicular = T \ y 90. 

Ex. 2. Here a— — 5, b= — 15, x 1 = 4, y'=5, and by 
placing these in the formula for the perpendicular, we find 

■p , -15-20-5 , -40 

Per. = ± — = — ■ = ± —— 

V5 2 + l V26 

=,', x40i/26 = ^± = 7.84+. 

(Page 118.) 

Ex. 3. Dividing both of the equations 
3y -f 5x = 4 
2y = Sx + 4, 
through by ?/ and transposing, they become 
y = — fu + i 
^ = f<23 4- 2. 
Here a= — f, a'=f, b = h U=2. Placing these values 
in the formula 

a;. = — ( r ), we get a;, = = — — ■ 

' \a-a<r & ' -f-| — V— f 

2x6 _ 4 
3x19 
Making the same substitutions in the formula 

dJh—aV __ a . fx|+| x2_2 + -V-_-V-_ 32 



1 9' 



V; = — i we get V; : 

To get the natural tangent of 30° we have 

Tan. 30o= ^g: = f^ = . 5773. 

cos. 30° .86603 

These values of x, , ^ , and tan. 30° placed in the equa- 
tion y — y { = tari. 30° (x—x{) give 
y-H = 0.-5778(^4- A). 



119] ANALYTICAL GEOMETRY. 

(Page 119.) 

Ex. 4. By transposing and divid- 2 

ing through by the coefficients of y, \ 

the equations 

2y— 3x= — 1 
2y + 3x=3. 
Become 

y-=\x— i. 

Whence a=f, a'— — f, Z>= — J 3 &'=§. 
These values put in the formulae 
l-U db-aV 



163 



D 



y 




\. 



X, = 



Vi 



Vi= 



a'— a 

The point C in which the lines intersect may then be con- 
structed. Since the first line (1, 1) ? intersects the axis of Y 
at the distance — j 3 and the second line (2, 2) intersects it at 
the distance f, the distance AB is 2 ; and because the lines 
(1, 1) and (2, 2) are equally inclined to the axis of X on 
opposite sides, they are also equally inclined to the axis of T 
on opposite sides ; hence, the triangle ABO is isosceles. The 

side AC=BC= VlHW= ^=i 1/I3=?i^±=l.20l4-. 

The perpendicular, CD, of the triangle is § ; hence the 
area of the triangle is =i x2=0. 66 + . 



(Page 119.) 
Ex. 5, The equations 

- 2\y + 3\x = - 2i, 

2f2/ - f« =4 

by transposing and dividing through by the coefficients of y 
become 

y = f jfc + J{, 

2/ = t 5 Was + f 



164 KEY TO [139 

Whence a = f f , a' = T \, b — |f , &' = f > 
substituting these in the formula for x { and y , we get 

45 5. 35 Q Y 17 K1 

nr — — ±* 2 — — — 1 32 — x ' — t>1 — 4Q 

and w,^^^- 4 — P-^ = ii— "-^ = 1.8. 

^ ' _5 3^ 5_ __ 1_3_0 

18 2 2 99" 

The form of the equation of a line passing through two 
given points is 

y—y ]] = y~ — Ul (x—x"), 
J u x u —x^ J ' 

and this will become the equation of the required line when 

we make 0^=0.49, ^ — 1.8, x n =3, y"=0. In this case the 

expression * „ ^ reduces to — 0.7171 + . 

x"—x l 

Placing these values in the above equation we get 

2/= -0.7171^ + 0.7171 x 3= -0.7171x+2.1523. 
And substituting the values of the two points in the 
formula. 



We have 



JD= |/(2.51) a +(1.8J 



CHAPTER II. 
THE CIRCLE. 

(Page 139.) 

Ex. 1. In all these examples the equation to be referred 
to is 

Comparing the equation a a -f-llx=80 with this, we have 



139, 187] AN ALYTICAL GEOMETRY. \Q§ 

e=ll ? ^!^ a =80; whence i? 2 = 160-f 121=281. B=V281 

A 

= 16.76, with which we proceed as explained in the text. 
Ex. 2. x*-3x-28. 

Here c=-S y ^^- = 28; whence iT= 56 + 9= 65, B 

A 

~V65 = 8.06. 
Ex. 3. x'~x^2. 
In this c=-l, -5^ = 2, B* = 5, B=V~5=2.23. 

Ex. 4. cc 2 -12x=-32. 

Inthise=-12 ? ^Z^-32, i2 2 =80, 5=8.94 

A 

Ex. 5. a 2 -12*=-36. 

This gives, c=-12 3 ^^=-36. 5 2 =72 ; 5=8.48. 

A 

Ex. 6. a 2 -12x=-38. 

Whence c=-12, ^!z^=_38, K 2 = 68, 5=8.24 

Ex. T. a 2 + 6x=-10. 

Whence c=6 ; ^Z^ = -10, 5 2 = 16, B=4. 



CHAPTER IV. 

THE PARABOLA. 

(Page 187.) 
Ex. 3. a?— t Vb=8. 

Comparing this with the equation 

5 2 -f-2&5=2c-Z> 2 
we see that 6 = — ^, 2c— 6 2 =8 ; whence 2c=8-h T i T =fff, 



166 KEY TO [187,223 

c=f£f=4+. We construct the pole by laying off the dis- 
tance 4 on the axis to the right, for the abscissa of the pole, 
and the distance — y 1 ,- for the ordinate. 

The values of x will be found to be +2.9+ and —2.7+. 

Ex. 4. %%* + %x= T \, or £ 2 + fa:=!!. 

Here 2b=±, 6=f, 2c-V=U, 2c=H + A, c = .504+. If 
the pole be constructed with these coordinates, the distances 
from the pole, above and below the axis, to the intersection 
of the perpendicular radius vector with the curve are .6 + 
and 1.4+ . 

Ex. 5. }tf-iy=2, or tf-%y=8. 

We find for this 26 = -|, 6=-J, 2c-Z> 2 = 8, 2c=8 + i, 
whence c=8.1 + . Constructing the pole with these coordi- 
nates, and drawing the perpendicular radius vector, we shall 
find the distances from the pole to the intersections with the 
curve to be =3.17+ and —2.5 + . 



CHAPTER VI. 

INTERPRETATION OF EQUATIONS. 

(Page 223.) 
Ex. 1. We find the abscissas of the vertices of the di- 
ameter whose equation is y=—x, by placing the quantity 
— 2x(x— 2) under the radical in the general value of y equal 
to zero ; that is we make 

— 2m(x-2)=0 
which gives the two values sc f =0 ; x ]] —% These values of x 
substituted in the equation of the diameter y=—x, give 
2/=0, y fl =—2 ; hence x' = 0, y'—0 are the coordinates of 
one vertex, and #"=2, ?/" = — 2 are those of the other. 



&36] ANALYTICAL GEOMETRY. 167 

The abscissa of the center x=l } and giving x this value in 
the proposed equation it reduces to 

Or, y' + 2y = l 

. Whence, y= — 1± V2=— 1± 1.41 +, 

Thatis #==+41 + and y= — 2.41 + . 

(Page 236.) 
Ex. 6. In the equation 

y*-2xy-x*-2y + 2x + 3 = 0, 
we see, by comparing it with the general equation that A = l, 
B——2, C= — l ; hence B 2 —4AC>0 and the equation re- 
presents an hyperbola. 

Solving the equation with reference to y, we find 

y=x + l±V2(x*— 1). 
Placing the quantity under the radical in this value of y, 
equal to zero we have x— + 1, and x= — 1 for the abscissas of 
the vertices of the diameter which bisects the chords parallel 
to the axis of Y ; hence the axis of Y bisects this diameter 
and is the conjugate to it. 

By making y equal to zero in the proposed equation it 
becomes 

x 2 — 2x=3 
Whence, x = 1±V3+1=1±2, x=-\-3, or —1. 

Ex. 7. Comparing the equation 

y*—2xy+2x*—2x + 4=0 
with the general equation, we find, 

A = l, B=-2, (7=2; therefore B 2 -4AC< 0, 
and the analytical condition for the ellipse is satisfied ; but 
since the equation can be put under the form 

(y- x y+(x-iy+3=o 

we see that all real values of x and y will make the first 



168 KEY TO [237 

number the sum of three positive quantities, and hence such 
values cannot satisfy the equation. 



(Page 237.) 

Ex. 8. 2/ 2 — 2xy+x*+x=0. 

This represents the parabola, because A = l, B——2, 0=1 
and, therefore, Z? 2 — 4^4(7— 0. The curve passes through the 
origin, since the equation contains no absolute term and is 
therefore satisfied by x=0, y = Q. 

Solving the equation in respect to y, we find 

y — x ± V— X, 

from which we conclude that all positive values of x will 
give imaginary values to y, or that the curve does not extend 
in the direction of x positive. 

If the equation be solved with reference to x we shall find 

x = -^- ± y -y> + J A u = -^— ±iVl—4y. 

■ The radical part of this value of x shows that any posi- 
tive value of y greater than }, will render x imaginary, but 
that x will be real for all negative values of y. Hence the 
curve extends indefinitely in the direction of x and y nega- 
tive. Substituting \ for y, in the equation, it becomes 



lx+x*+x=0 



Or, x- + ±x+ T \ = (x+iy = 

Whence, ic-fi=0, x= — }. 

A line drawn through the point whose coordinates are 
y= + J, x= — I, parallel to the axis of X, will limit the 
curve in the direction of y positive ; but it will be unlimited 
in the direction of y negative. 

Ex. 9. tf-2xy + x 2 -2y-l = 0. 

The curve represented by this equation is the parabola. 



237] ANALYTICAL GEOMETRY. 169 

Since we have for it, A—\, B=—2, 0=1, and, therefore, 
B 2 -4AO=0. 
Transposing and factoring it becomes 

Whence, 



y=x + l±V — sc a + l+rc 2 + 2;c + l =x + l±V2(x + l). 

Also, £— 2xy= — y 2 + 2y + l, 

Whence, x=y± V2y + 1. 

The value of y shows that all negative values of x greater, 
numerically, than —1 will render y imaginary, and the value 
of x shows that all negative values of y greater, numerically, 
than —\ will render x imaginary. The curve is therefore 
limited in the direction of the negative axes ; but it is un- 
limited in the opposite direction. 

Ex. W. y*—4xy + 4x' 1 =0. 

Since in this equation, A=l, B=—4, 0=4, we have 
B 9 —4AO=0 which is the condition for the parabola. But 
we have also the further conditions (page 219), 
BD-2AE=Q, D*-4AF=0. 
In consequence of the first of these conditions, the parabola 
would reduce to two parallel straight lines, and in consequence 
of the second, the two straight lines reduce to one, or coincide. 
The first member of the equation is a perfect square, and 
we therefore have 

(y—2x) 2 = 0, "or y=2x. 

Hence the line to which the parabola reduces, passes 

through the origin of coordinates and makes, with the axis 

of X, an angle, having 2 for its natural tangent. The degrees 

and minutes of the angle are found thus 

Log. 2 = 0.301030 

Log. K = 10.000000 



Log. tangent 63° 26', 10.301030. 



170 KEY TO [237,943 

The complement of 63° 26' is 26° 34', which is the angle 
the line makes with the axis of Y. 



Ex. 11. tf-2xy + 2x*-2y+2x=0. 
Here A = l, B=-2, C=2. Hence B'-4AC<0, and the 
curve is an ellipse. 

Solving the equation with reference to y, we get 

y=x+l±V— 05 9 + 1. 
Placing the quantity under the radical sign equal to zero, 
to get the abscissas of the vertices of the diameter whose 
equation is y=x + l, we find 

#=±1, or x=-\-l and x—— 1. 
The value of ^ shows that all substitutions for a;, numeri- 
cally greater than 1, positive or negative, will render y 
imaginary. Hence parallels to the axis of Y } drawn at a 
unit's distance from it on either side, will limit the curve. 



CHAPTER VII." 

ON THE INTERSECTION OF LINES AND THE GEO- 
METRICAL SOLUTION OF EQUATIONS. 

(Page 243.) 

Ex. 3. Given tf— 48y= 128, to find the values of y by 
construction. 

Making y=nz and substituting in the given equation it 
becomes 

»V— 48nz = 128, 

m 3 48 128 

Whence, a 3 — — z — — - . 

7 n n 3 



243] 



ANALYTICAL GEOMETRY. 



171 



If now, we assume n—4, this equation will red ace to 
z s -3z = 2, 
and the construction of the values of z is the same as in 
example 2 ; that is 3=2, 3=— 1, 3= — 1, and since y=nz, 
and n=4 we have, 

Ans. ?/=+8, —4, —4. 




Ex. 4. Given y"~ 13?/=— 12 ; to find the values of y by 
construction. 

Comparing this equation with equation ( G ) page 241, 
we have 

4-4a=-13, 85= -12, 

Whence, a=4£, h——\\. 

Constructing the parabola 
with the distance from the 
focus to the directix for the 
unit, we then lay off on its 
axis, in the positive direction, 
the distance AD = 4^, then 
BC——\\ will determine the center of the circle. The cir- 
cumference described with C as a center and CA as a radius, 
cuts the parabola in the points m, m', ra", which are at the 
distances from the axis of the parabola -fl, +3, and —4, 

respectively ; hence 

Ans. y=-\-l, +3, —4. 

The different values of x, on page 246, corresponding to 

the assumed values 30.0388, 25 —5, &c, of y are thus 

found ; 

Equation {A) page 245, when y— 30.0388 becomes 
a 3 = 13a? + 12 = 30.0388, 
■ Or, x* —13a =18.0388. 

Whence, 4-4a=-13, 86 = 18.0388 (see eq. (G) page 241). 

Therefore, a =4}, 6 = 2.25485. 



172 KEY TO [^@9 

These values of a and b will give us the center of the cir- 
cumference, the intersections of which with the parabola, 
determine the values of x. 
So when y=25 we shall have 

a; 3 -13a;=13 
4-4a = — 13, 85 = 13; a=4j, #=lf, 
and so on, the value of a being constantly equal to 4£. 



CHAPTERS VIII., and IX. 

STRAIGHT LINES IN SPACE AND PLANES. 

(Page 269.) 
Ex. 1. What is the distance between two points in space 
of which the coordinates are 

03=1, y=—5, z=— 3; #'=4, ?/=— 4, «'=1. 
Substituting these values of x, y, z, x\ y' y z\ in the 
formula 

D*=(x-x>y + (y-y'Y + (z-zy 

found on page 254 (Prop. 6), and taking the square root we 

get 

D=V{s+2y+(5+iy+(-2-6y 



D=V25 + 36 + 64 = V125 == 11.180 + . Ans. 

Ex. 2. Of which the coordinates are 

x—1, y— — 5, z=— 3; a;' =4, y , = — 4, a=l. 

As in Example 1, substituting these values of the coordi- 
nates of the two points we rind 

2>= l / (l-4) 2 + (-5 + 4) 2 + (-3-l) 2 

-V 9+1+16= ^26 

= 5.098+ = 5 A- nearly. ^rcs. 



9©9] ANALYTICAL GEOMETRY. 173 

Ex. 3. The equations of the projections of a straight 
line on the coordinate planes (xz), (yz), are 

x=2z+l, y=\z-2, 
required the equation of the projection on the plane (xy). 
Multiplying the second equation through by 6, we have 
6y=2z-l2 
which makes the coefficients of z the same in the two equa- 
tions. Subtracting the first equation from this last, member 

from member, we get 

6y— x=— 13 

Or, y={x—2}. Ans. 

Ex. 4. The equations of the projections of a line on the 
coordinate planes {xy) and (yz), are 

2y=x—5 and 2y = z— 4. 
Eequired the equation of the projection on the plane (xz). 
By subtracting the second of these equations from the first, 
member from member, we have 

0=x—z—l, or 05=3+1. Ans. 

Ex. 5» Eequired the equations of the three projections 
of a straight line, which passes through two points, whose 
coordinates are 

x j =2, y> = l } sj' = 0; and x"=-3, y"=0, ss" = — L 

Placing these values of x ] , x u , y', &c, in the formulas on 
page 252 (Prop. 8), which are 



the first becomes 



x—x ] — -j. Xz— z 1 ) : y—y ] — y — — ±-{z—z\ 



3 2 

x— 2 = — — z, or x=5z + 2, 

and the second ) Answers. 

y-l=^z, or y = 2+1, 



174 KEY TO [2©9 

which are the equations of the projections of the required line 

on the planes (xz), (yz). 

We eliminate z from these equations by subtracting the first 

from the second after multiplying the second through by 5 : 

Thus 

5y — 5z + 5 

x = 5z + 2 



Subtract and 5y—x=5—2, or 5y—x + 3. Ans. 

Which is the equation of the projection of the line on the 
plane (xy). 



x — z + 2 ] 
of the 1st ; and < _, > of the 2d. 

'=-3 + 1 J 



Ex. 6. Kequired the angle included between two lines 
whose equations are 

„ > of the 1st ; and < 
y=2z + 6j \y. 

Keferring to the formula 

~ Tr l + aa' + 55' 

Cos. r = - — — - — 

found on page 257, Prop. 8, and comparing the coefficients 
of z in the given equations with those of the equations in 
that proposition, we have 

a=% a' = l ; b=2, b' = -l. 

These values substituted in the formula give 

n jt 1 + 3-2 2 , _ 

Cos. V= f = -— = a-Vi/42 

1/1 + 9+4^1 + 1 + 1 \%2 

l/42 = 6.4808 ; hence i^M = .308609. 

That is 

Cos. V = .308609. 

The table gives .30846 for the cosine of 72° 2', and .30874 
for the cosine of 72° 1' ; hence we have the proposition 
28 : 13 : : 60" : x" = 28", 
And T=72° 1' 28". Ans. 



*1©9] ANALYTICAL GEOMETRY. I75 

Ex. 7. Find the angles made by the lines, designated in 
the preceding example, with the coordinate axes. 
Formulas (5), (6) and (7), page 255, Prop. (7), are 



Cos. X- 

±Vl + a 2 + b 2 


(5) 


Cos Y— h 


(6) 


±Vl + a 2 + b* 


Cos. Z- 1 


(7) 



±l/l + a 2 + Z> 2 

In these, to find the angles the first line makes with the 
coordinate axes, we must make a=3, b—2. 
These values placed in the above formulas give 

Cos. X= — = — 3 — 

±4/1+9+4 ±Vu 

= ^li = T V1/126 = T V(11.224+). 
Cos. X = .80179. By the table, X = 36° 42'. ,4ws. 

Cos. Y= —*— = T \ V '5Q = T V(7.4833 + ) 

±1/14 

Cos. F= .53452. y = 57° 41' 20". <4ws. 

C0S ' Z = ^7H = tV^H - tV(3.7416 + ) 

Cos. Z = .26727. s = 74° 29' 54". -4ws. 

In like manner to find the angles that the second line 
makes with the coordinate axes, we must make a=a' = l, 
&=&'=— 1. Formulas (5) and (7) then give 

Cos. X= cos. Z = -L = 1 1; 3 = 1(1.73205) 

Cos. X= cos. # = .57735, X=^=54° 44'. Ans. 

Since &'=— 1, the cosine of Tmust be taken with a minus 
sign, if the cosines of X and Z are taken with the plus sign, 



176 KEY TO [270 

and the converse. But these cosines have the same numerical 
value ; therefore Y is the supplement of 54° 44' ; that is 
T= 125° 16'. Ans. 

Ex. 8. Having given the equations of two straight lines 
in space, as 

X = ? + l\ of the 1st ; and j * = * + \ X of the 2d, 

to find the value of 0', so that the lines shall actually inter- 
sect, and to find the coordinates of the point of intersection. 
If the equation 

a _ a) _ — 0' 
a — a b — b' 
found on page 252, Prop. 4, is satisfied by the constants in 
the equations of any two straight lines, such lines will inter- 
sect. Any five of these constants being given, the equation 
will determine the sixth. In the present example we have 
x—1, x , =2 } a=3, a'=l, b=2, Z>' = — 1, (3=6, from which to 
find 0'. 

Placing these values in the above equation it becomes 
1-2 _ 6 - 0' 
3^~T _ 2~+T J 
Whence 20' = 15; 0' = 7|. ^ws. 

The formulas for the values of x and y, the coordinates of 
the point of intersection given on the same page, are 
aa — aa) b& — b(3' 



we get 



x --aZ7ar> y~ b _ h , 



6-1 01 , 15 + 6 „ 

x = — = 2i, and y = — g— = 7. 



ft /*, 



The value of 2 may be found from any one of the given 
equations, by substituting for x or y, the value just found. 



270] ANALYTICAL GEOMETRY. 177 

Take for example the 1st, £=3z + l ; it becomes 2^=33 
— 1 ; whence 6z= 3, b=£. 
Hence the 



Ans. 



f (3<=7h y=7. 
\ x=2\, »=t 



Ex. 9. Given the equation of a plane 
8x=3y+z— 4=0, 
to find the points in which it cuts the three axes, and the 
per]3endicular distance from the origin to the plane. 

The point in which the axis of X pierces the plane is 
found by making y=0, 2=0, in the equation, by which we 
get 8#— 4=0, or x=±. Similarly by making x=0 } 2=0, 
we have y— — H ; and x—0, y=0 } gives 2=4. 
Formula (7), page 276, which is 
_ D 

P ~ VA' + F + C*' 
will give the perpendicular distance from the origin to the 
plane by making A— 8, B = —3, (7=1, D=— 4. 
These values give us 

._ -4 = _4_ = 4^74 

P " ±4/64+9"+! ~ V74 74 



•1184 = 34409 = ^ 



74 

( ^=.4i 



Hence the . ( aj.=i, «/,= — H, 3, =4. 



4649 + . 



Ex. 10. Find the equations for the intersections of the 
two planes, 

3a— 4^/ + 23-1=0, 
7#— 3y— 3 +5=0. 

If 2 be eliminated between these two equations the result- 
ing equation will be the equation of the projection of the 



178 KEY TO [269 

intersection on the plane (xy). We eliminate by multiplying 
the second equation through by 2 and then adding it to the 
first, member to member ; thus 

3x— 4y + 2z— 1=0 

Add 14x—6y—2z + 10=0 

17x— lOy + 9=o7~ 1st Ans. 

We eliminate y from the given equations by multiplying 
the first through by 3, and the second by 4, and subtracting 
the first result from the second. 

Thus, 28x-12y-4z + 2Q=Q 

Subtract, g x ^i2y+6z— 3=0 

19a—10B + 23=0. 2d Ans. 

We find the angle included between the two planes by the 
formula 

AA< + BB< + CC 



Cos. V = 



fa "I 



VA*+B* + C< VA<* + B> 2 + C< q 
in which we must make ^4 = 3, B=— 4, 0=2, A 1 — 7, B j =— 3, 
(7' =— 1. These substitutions will give 

21 + 12-2 31 31 



Cos. V= 



V9 + 16+4i / 49 + 9 + l 


V29V59 


31V 1711 V 1644271 
1711 1711 


1282.24 
1711 " 



^1711 



Hence, Cos. F=.74940 ; F=41° 27' 41". Ans. 

Ex. 12. The equations of a line in space are 

x=— 2z+l, and ?/=33+2. 
Find the inclination of this line to the plane represented 
by the equation 

8^-3?/+2-4=0 
The formula to be used in this example is that found on 
page 268, Prop. 7, which is 

Aa+Bb + C 
Vl+a' + b^C+B'+A* 



371] ANALYTICAL GEOMETRY. 179 

In this we must make a—— 2, b — 3, A=S, B=—3, 
C=l, which will give 

-16-9 + 1 24 12 



Sin. v = 



|/i + 4 + 9i/l + 64+9 VUV74 V7 V%7 



1/37296 193.147+ _ 74571 



259 259 

Whence v = 48° 13' 13". ^ras. 

Ex. 13. Find the angles made by the plane whose equa- 
tion is 

8x— 3?/ + 2-4=0 

with the coordinate planes. 

The formulas to be used in this case are 

COS. (#Z/) = - , COS. (OJZ) r- 



V^ 2 + 5 2 + (7 2 ' V^ 2 +£ 2 + (? 2 ' 

cos. te) = — — — , and are found on page 264, Prop. 3. 

■ VA' + JF + C' . 

By comparing the given equation with the general equa- 
tion of the plane, we find, A =8, B=—3, (7=1. 

These values being substituted in the first of the above 
formulas it becomes 

Cos. {xy)=-=^== = ^= = T VV74 
V64+9 + 1 ^74 

Cos. (a;y) = T V(8.6023) = .11625. 

The Table gives 83° 19' 27" for the angle whose natural 

cosine is this decimal. 

Similarly the second formula gives 

+ 3 
Cos. (xz)= -7=g=- tV 4/666 = - .34874, 

and the third 

i o 

Cos - 00 = ~j=i = T V 4/4736 = .92998, 



180 KEY TO [271 

which correspond respectively to the angles 110° 24' 38" and 
21° 34' 5". 

83° 19' 27" with the plane (xy). 
110° 24' 88" " " ", (?cz). 

21° 34' 5" " " " (yz). 



Hence the 

Ans. 



Ex. 15. Find the equation of the plane which will cut 
the axis of Z at 3, the axis of X at 4, and the axis of Y at 5. 

The equation of the required plane must have the form 

Ax+By+Cz+D=0 ? 

and such values of the coefficients A, B and C, are to be 

found as will cause the plane represented by the equation to 

cut the coordinate axes at the specified points. 

Keferring to the Scholium on page 260, Ave find the follow- 
ing expressions for the distances from the origin to the points 
in which a plane cuts the axes, viz. : 

If in these we give to x, y and z, their values by the con- 
ditions of the problem, we have 

6 ~ C> b ~~B> 4_ A' 

Whence, G=-~, A=-^, B=-~. 

' 3' 4 ? 5 

Substituting these values in the general equation of the 
plane it becomes 

Or, 15x + 12y + 20* = 60, 

by multiplying through by — — ; and by dividing this last 

equation through by 3 we get finally. 

Ans. 5x +4ty + 6 \z— 20. 



£72] ANALYTICAL GEOMETRY. 131 

Ex. 16. Find the equation of the plane which will cut 
the axis of X at 3, the axis of Z at 5, and which will pass 
at the perpendicular distance 2 from the origin. At what 
distance from the origin, will this plane cut the axis of Y? 

Let PQB be the required plane cut- 
ting the axis of X at P, making OP =3, 
and the axis of Z at B making OR =5. 
Op is the perpendicular let fall from 
the origin upon the plane, and by con- 
dition it is equal to 2. 
% Conceive a plane to be passed through 

the axis of Z, and the perpendicular Op intersecting the 
required plane in the line BS, and the plane (xy) in OS. 
Since the plane through OB and Op is perpendicular to both 
the planes (xy) and PQB, it is perpendicular to their inter- 
section, PQ. In the figure we therefore have the right angled 
triangles BOp, BOS, BOP. BPS, OPS and OPQ. 

From the triangle BOp, we have 




Bp = ^/B0 2 -Op = 4- 7 5 2 -2 2 = V21, 
and from the triangles PiOp, BOS, 

Bp : OB : : OB : BS } 

Or, V21:5 : : 5 :BS = -^L. 

V21 

The triangle BOP gives 



BP=^/BO*+P0 2 = |/5 a + 3*=i/34, 



Whence, PS= ^/BP 2 -EST = |/34— W-= Vff. 



182 KEY TO [272 

Then from the similar right angled triangles OPS, OPQ, 
we have 

PS : OP : : OP : PQ 

9 9i 7 21 



Or, V\\ : 3 : : 3 : PQ 

Whence, OQ=/f^^=l/^-9 =\/ m 
, which may have either sign. 



*!L-9=, 

89 f 89 



|/89 

The equation of the plane will be what the equation 

Ax+By+Ck+D=Q 

becomes, when we substitute in it the values of A, B and (7, 
given by the equations 

r __D _ _D _30 _ _D 
°~~ C : A> 4 /89~ B' 

We thus get 

D V89D D 

Or, lOx + V89 y + 6z-30 = 0, 

which is the equation of the required plane, and it cuts the 

30 
axis of Y at the distance 00=—=, from the origin. 

V8d 

Ex. 17. Find the equations of the intersection of the 
two planes whose equations are 

3x-27j-z-4=0, 
7x-\-3y+z—2=0. 

By adding these equations, member to member, we elimi- 
nate z and get for our result 

10jc + 2/-6=0 
which is the equation of the projection of the intersection of 
the two planes on the plane (xy). 



373] ANALYTICAL GEOMETRY. 183 

Multiplying the same equations through, the first by 3, 
and the second by 2, we have 

9x-6y-3z-12=0, 

Add, 14x + 6y + 2z- 4—0 

And our result 23x — z — 16 = 0, is the equation of the 
projection of the intersection on the plane (xz). 

In like manner multiplying the first through by 7, and the 
second by 3, we find 

21x—Uy-7z—28=0, 

Subtract, 21a + 9y + 3z— 6 = 0, 

And our result, 23?/ + 102 + 22=0, is the equation 

of the projection of the intersection on the plane (yz). 

Ex. 18. Find the inclination of the planes whose equa- 
tions are expressed in Example 17. 
The equations of the planes are 

3x-2y-z-4=0, 
7x + 3y+z-2 = 0. 
The formula for the cosine of the inclination of these two 
planes is 

Co, V- U+B* + O0> 



VA'+B^+CWA^+B^ + O* 
found on page 267, Prop 6. In this we must make the fol- 
lowing substitutions, viz. : 

A=3, B=-2, <7=-l; A' =7, B'=S, C=+l, 
by which we get 

21-6-1 14 VTi 



Cos. V= 



4/9+4+I1/49 + 9 + I V14V59 V59' 



m n JT 4/14 1/141/59 1/826 Aamn 

Whence, Cos. V= = — -- — = — — - = .48712, 

5 1/59 59 59 ? 

and by the Table this is found to correspond to the angle 
60° 50' 55 n } or the supplementary angle, 119° 9' 5". Ana, 



184 



KEY TO 



Ex. 19. A plane intersects the coordinate plane (xz), at 
an inclination of 50°, and the coordinate plane (yz), at an 
inclination of 84°. At what angle will this plane intersect 
the plane (xy) ? 

We employ in this case the formula 

Cos. 2 (^) + cos. 2 (a:z)-f cos. 2 (2/z)=l,. 
which is found on page 264. From it Ave get 

Cos. '{xy) = 1 — cos. 2 (a^) — cos*(yz). 
But cos.(o3) = cos.50° = .64279, and cos.(?/2) = cos.84 = . 10453. 
Whence, Cos. 2 (^) = l-.4241054050-.0109265209 

Cos. (a^) = .i/.5758945950 = .75887, 
which "by the Table corresponds to the angle 40° 38' 6". Ans. 



MISCELLANEOUS PEOBLEMS. 



Ex. L The greatest or major axis of an ellipse is 40 feet, 
and a line drawn from the center, making an angle of 36° 
with the major axis and terminating in the ellipse, is 18 feet 
long ; required the minor axis of this ellipse, its area and 
eccentricity. 

Suppose AB to be the major axis, 
C the center, and F the focus of the 
ellipse, and let Cp be the line drawn 
from the center making an angle of 
36° with AB, and terminating in the 
curve at p. From p let fall the per- 
pendicular pD on the axis, and pro- 
duce this perpendicular to meet the circumference described 
on AB, as a diameter in P, and draw CP. 




273] ANALYTICAL GEOMETRY. 185 

Then in the right angled triangle CpD we have 
1 : sin. pCD : : Cp : pB = Cp.sm. pCB. 
1 : cos. pCD : : Op : OT = Cp.cos. pCD. 

In the Table we find sin. 36° = .58779, cos. 36° = .80902, 

Hence, pD=l% x .58779=10.58022, 

And GD=18 x .80902 = 14.56236. 

From the right angled triangle CPD, we get 

PD=^/CP 2 -(W*= V 400-212.062439 

= 4/187.937561 = 13.709. 

Now we have an ordinate of the ellipse drawn to its trans- 
verse axis, and the corresponding ordinate of the circle de- 
scribed on that axis. Calling the semi- transverse axis A, and 
the semi-conjugate axis B, we have, 
A:B::PD:pB, 

That is 20 : B : : 13.709 : 10.58022. 

Whence, B = ^^^ = 15.4376, 

13.709 

And 2B = 30.8752. 

The area of an ellipse is measured by the product of its 

semi-axes multiplied by 3.14159 + . 

Hence for the required ellipse, we have 

Area =ttAB=3. 14159 x 20 x 15.4376 = 969.972 + . 

Denoting the eccentricity by E, we have 

VA'-B* ^400-238.3195 12.715 „ OWTK 
E= —j— = — = __ = .63575. 

Collecting our results we therefore have 

Minor axis, 30.8752 feet. 
Ans. ^ Area, 969.972 sq. feet. 

Eccentricity, .63575 feet. 

In the second miscellaneous problem it was assumed that 
if the side of an equilateral triangle be denoted by a, the 



186 KEY TO ANALYTICAL GEOMETRY. [273 

line drawn from the center to the vertex of either angle of 

the triangle, will be represented by — =. To prove this, let 

V 3 

C AB be an equilateral triangle of 

which D is the center, by which we 

mean the center of the inscribed or 

circumscribed circle. Draw AD 

and CD, producing the latter to 

meet the base of the triangle in E. 

Now since the angle CAE is 60° 

and AD bisects this angle, the angle DAE is 30°, and DE 

is therefore equal to one half of AD. Denote AB by a and 

DE by x; then AD = 2x. 

In the right angled triangle ADE, we have 




Or, 

Hence, 



AE +DE' 

4-x 2 = 4x 2 , 



3x 2 =- 



a 



2|/3 



2x = AD = 



V3 7 



which was to be proved. 



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